CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE …trentschirmer.com/Intermediate_Algebra_Notes.pdf · CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 1. The basic laws

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATEGEOMETRY

1. The basic laws of Algebra

Section 1 of these notes covers the same material as Sections 1.1, 1.2, and 1.6 of thetext (in a somewhat very different way). If you need more details about a topic discussedhere, ask me or look there.

1.1. The most important basic fact of algebra. This class is all about the manipu-lation of algebraic formulas. An algebraic formula is just a bunch of numbers, arithmeticsymbols, and letters of the alphabet thrown together, such as 2x2− 3 or 3xy√

2x. The letters

of the alphabet that occur in an algebraic formula are called unknowns, because theystand for hidden numbers that we cannot see.

Underneath it all, an algebraic formula stands for some hidden number. You can andshould, therefore, treat it like a number in all of your dealings with it. For example, justlike you can cancel the 4’s to show that

4 · 34

= 3,

you can also show that

(x2 + 1) ·√x

(x2 + 1)=√x

by canceling the (x2 +1)’s. However, just because a cancellation rule works for one choiceof numbers, doesn’t mean it will work for an algebraic formula

The Most Important Basic Fact of Algebra. When you simplify an algebraic ex-pression, whatever you do to it needs to work for every number you can plug in for itsunknowns.

So, for example, if you changed (x+3)2 into x2 +32 on a quiz, I would mark you wrong,because if you plug in x = 1 into the equation (x+ 3)2 = x2 + 32 you get

(1 + 3)2 = 12 + 32,

which is false because the left side is 42 = 16, while the right side is 1 + 9 = 10.Usually, if you are not sure whether or not one of your algebraic moves is right, that

means it is wrong. In situations where you are unsure, you should always check yourformula by plugging in actual numbers for the unknowns and seeing if they work.

Example 1.1. Plug some numbers into the following equations to see if they work.

• (x+ y)3 = x3 + y3

Solution: Plugging in x = 1 and y = 1 we get1

CLASS NOTES: INTERMEDIATE ALGEBRA AND COORDINATE GEOMETRY 2

(1 + 1)3 = 13 + 13

23 = 1 + 1

8 = 1!!!

(1.1)

This last equation is clearly false, so the formula (x + y)3 = x3 + y3 is false aswell.• x2+2x+1

x+1= x+ 1

Solution: With x = 3 we get:

32 + 2 · 3 + 1

3 + 1= 3 + 1

9 + 6 + 1

4= 4

16

4= 4

(1.2)

Which checks out, and if you plug a few more numbers in yourself you’ll seethat it keeps on working. This indicates that the equation is in fact true, althoughto actually know for sure we need to use some laws of arithmetic.

IMPORTANT: In this last example, plugging in x = −1 yields the following:

(−1)2 + 2(−1) + 1

(−1) + 1= (−1) + 1

1− 2 + 1

0= 0

0

0= 0!!!

(1.3)

This last equation is a problem, but not because it is false. Rather, it is senseless,because we divided by 0 on the left side. In this case, we don’t say that the algebraicformula is wrong. Instead, we just say that it doesn’t make sense for x = −1 and moveon.

1.2. Laws of algebra from laws of numbers. You are expected to be familiar with thebasic algebra discussed in Sections 1.2 of the text, here is a small, representative selectionof problems for you to test yourself with.

Homework 1, # 1 Compute the following expressions.

(a) 710

+(−4

5

)(b) −4.3− (−8.7)(c) (−4)(−2)3(−2)(d) (−3)2

(e) −32

(f)(32

)2(g)

(−3

2

)3


(h) (5−6)2−2(3−7)89−3·52

(i) −7(2x+ 3)(j) 7− 4[3− (4y − 5)]

Look at problems 13-130 in Section 1.2 for more practice.

The laws of arithmetic are facts and equations that are true for all numbers. You useda bunch of them when you were doing the problem set above. Since they are true for allnumbers, we can also plug algebraic expressions into them and get out true statements.There are too many of them to list here, however here are three of the ones I see studentshave trouble with most often when using them on algebraic expressions.

The Distributive Law: The distributive law says that, for any three numbers a, b, andc:

(1.4) a(b+ c) = ab+ ac

So, for example, equation 1.4 says that 5(2 + 3) = 5 · 2 + 5 · 3, which checks out:

5(2 + 3) = 5 · 2 + 5 · 35 · 5 = 10 + 15

25 = 25

(1.5)

Since 1.4 is true for all numbers, we can also apply it to algebraic formulas, for example:

(x2 + 1)(3x− 1

x) = (x2 + 1)3x− x2 + 1

x.

Here I just substituted x2 + 1 for a, 3x for b, and 1x

for c in 1.4. Notice the distributivelaw can also be applied backwards to “pull out a common factor” like so:

3x3 + 6x2 = 3x2x+ (3x2)2 = 3x2(x+ 2).

Here the distributive law 1.4 is applied in the second equality by setting a = 3x2, b = x,and c = 2. However, notice that what makes the “reverse” distributive law difficult to useis the first step, where I had to recognize the common factor 3x2. This just takes practiceto get good at.

Exercises 111-130 in Chapter 1.2 of the book provide good practice for the distributiveproperty. Try these ones right now:

Homework 1, # 2

(a) Use the distributive property to rewrite −7x(2x2 + 1).(b) Use the “reverse” distributive property to factor 8x2 − 4x3.

Exponents: Remember if n is a positive whole number, then by definition

an =

n︷︸︸︷a · a · · · a .


We used this formula on the last list of problems. Recall at the beginning of Section 1.1,we showed that the equation

(x+ y)3 = x3 + y3

is false. To find the correct simplification, we use the exponential law with a = x + y,then simplify by using the distributive law:

(x+ y)3 = (x+ y) · (x+ y) · (x+ y) = (x2 + 2xy + y2)(x+ y) = x3 + 3x2y + 3xy2 + y3.

Also remember that negative exponents are defined by the formula

a−n =1

an.

This definition is made to ensure that that the rule anam = an+m works even when n andm are not both positive whole numbers. The other rules you need are

(ab)n = anbn

and

(an)m = anm.

Remember these rules work for any numbers n,m, even negative ones. Here are someproblems from section 1.6 that will check how good you are with exponents.

Homework 1, # 3 Simplify the following:

(a) (2xy2)3

(b)(x3

y−4

)−4(c) 30x2y5

−6x8y−3

(d) (−4a2)2(2a−5)

Look at problems 1-124 in Section 1.6 for more practice

Fraction addition: The “easy” law of fraction addition tells us that, for any threenumbers a, b, and c:

(1.6)a

c+b

c=a+ b

c

This allows us to perform the following kind of simplification:

x2 + 2x

x=x2

x+

2x

x= x+ 2.

Here the first equality is just Equation 1.6 in reverse with a = x2, b = 2x, and c = x, andthe second is basic cancellation.

The problem with the easy law of fraction addition is that you need the bottom of bothfractions to be the same in order to add them together. When your fractions are differenton the bottom, you need to use the “hard” law of fraction addition:

(1.7)a

b+c

d=ad+ bc

bd


For example:

x+ 1

x− 1+

x

x2 − 1=

(x+ 1)(x2 − 1) + x(x− 1)

(x− 1)(x2 − 1)

Where here I just plugged a = x + 1, b = x − 1, c = x, and d = x2 − 1 into Equation1.7. You should also be able to multiply the top and bottom out to get

x3 + 2x2 − 2x− 1

x3 − x2 − x+ 1.

By the end of this class, you will have even deeper skills which allow you to see that thislast expression can be simplified down further to

x2 + 3x+ 1

x2 − 1

We won’t go further into the nitty gritty of adding fractions of algebraic expressionsuntil much later in the class, so there’s no need to brush up further on them right now.But here’s an example of the kind of thing you will need to be comfortable doing:

Example 1.2. Rewrite the expression (x + 1)−1 + (x − 1)2 as a single fraction with noparentheses.

Solution: This requires use to use all three of the major arithmetic laws I pointed out(and some minor ones that I didn’t).

(x+1)−1+(x−1)2 =1

x+ 1+

(x− 1)(x− 1)

1=

1 · 1 + (x+ 1)(x− 1)(x− 1)

1 · (x+ 1)=x3 − x2 − x+ 2

x+ 1.

2. Solving for x, using algebra in real life.

This section discusses Chapters 1.4 and 1.5 of the book.

2.1. Solving for x. In Section 1 of these notes, we went over some ways to decide whetheror not an equation f(x) = g(x) is true for all values of x, so that we could tell whetheror not it was okay to change f(x) into g(x).

This task is different from solving a given equation f(x) = g(x) for x. In this case, wedon’t expect f(x) to be equal to g(x) for all values of x. We just hope they are equal forsome value of x, and want to find it.

Example 2.1. Solve the equation 3x− 1 = 2x+ 1 for x.

Solution: In addition to substituting equivalent algebraic substitutions for one another,we can do almost anything we want to one side of the equation, so long as we do the same


thing to the other side.

3x− 1 = 2x+ 1

3x− 1 + 1 = 2x+ 1 + 1

3x = 2x+ 2

3x− 2x = 2x+ 2− 2x

x = 2

(2.1)

So we see that this equation does have exactly one solution, x = 2.

Homework 2, #1: Solve the following equations for x.

(a) 7x+ 4 = x+ 16(b) 3(x− 4)− 4(x− 3) = x+ 3− (x− 2)(c) 5 + x−2

3= x+3

8

(d) 8x+ 1 = x+ 43(e) 7(x+ 1) = 4[x− (3− x)](f) x+1

3= 5− x+2

7

For further practice, look at problems 1-38 of Chapter 1.4 of the book. We will learnmore advanced x-solving techniques later in the course.

2.2. Algebra in real life: In my experience, word problems are just learned best byworking examples, although the book breaks down the process into 4 steps in Chapter1.5, which you may find helpful.

Often, word problems in math classes are somewhat contrived, but I happen to like thisone from the book (it is a slight rewording of Example 3 of Chapter 1.5):

Example 2.2. A store offers you a choice of two texting plans. Plan A has a monthlyfee of $20 with an additional charge of $0.05 per text. Plan B has a monthly fee of $10with an additional charge of $0.10 per text. Which one should you get?

Solution: As I’ve worded the problem, it doesn’t have a right or wrong answer, just likea real world problem will not have a right or wrong answer. But we can still use math tomake a more informed decision.

The store has offered you a very common kind of deal. Plan A requires you to pay moreup front than Plan B, but you only have to pay half as much money per message withplan A. So Plan B will be cheaper if you don’t plan on texting very much, and Plan Awill be cheaper if you plan on texting a lot. Only you can guess how many text messagesyou will send in a month, but using math we can at least figure out which plan will becheaper based on this guess.

So, suppose x is the number of text messages you think you will send in a month. ThenPlan A will charge you the $20 up front, plus an additional $0.05×x at the end of themonth. So your total Plan A expenses will be 20 + .05x dollars.

Meanwhile Plan B charges you only $10 up front, but then it tacks on $0.10 per messageyou send, which puts your Plan B expenses at 10 + 0.10x.

As we observed earlier, Plan B will be cheaper initially, but once the number x of textmessages gets big enough, Plan A will become cheaper. In order to figure out how big x


needs to get in order to make Plan A cheaper than Plan B, we must find the value of xwhere Plan A and Plan B cost exactly the same. But given the formulas above, this justmeans solving the equation

20 + .05x = 10 + .1x

for x. We get:

20 + .05x = 10 + .1x

10 + .05x = .1x

10 = .05x

200 = x

(2.2)

Therefore, even if you are tempted to go with Plan B because it’s cheaper at first, PlanA will be the better choice if you send more than 200 texts in a month. And since 200text messages comes out to just 200

30≈ 6.67 text messages a day during the average month,

I’m willing to bet that most of us here would be better off with Plan A. Incidentally, thisis usually the way things work in the world: those who choose to pay more up front payless in the long run.

Homework 2 #2:

(a) If the temperature is C degrees Celsius, then it is F = 95C + 32 degrees Farenheit.

Suppose it is a 98 degrees Farenheit in Salinas. What’s the Celsius temperatureof Salinas?

(b) An Inn charges you $250 to stay the night, and this price includes an 8% salestax. How much money goes to the Inn, and how much to the government?

(c) The toll for a bridge is $5.00 every time you cross it, unless you purchase a discountpass for $30, which lowers the price to only $3.50 every time you cross it, for amonth. How many times a month would you have to cross the bridge in order tomake the discount pass worth it? Please show your work using algebra.

(d) After a 20% reduction, you purchase a television for $336. What was the televi-sion’s cost before the reduction?

Problems 17-40 in Chapter 1.5 of the book are all great for further study.

3. Functions and Graphs

In this part of the notes we cover Section 2.1, then jump back to 1.3, and end with alittle bit of 2.2.

3.1. A conceptual shift. The word set in the definition below is just a short mathe-matical word for a collection of objects.

Definition 3.1. If A and B are sets, then a function f from A to B is a rule whichassigns a single member of B to every member of A.

This definition is quite general, and Chapter 2.1 of the book gives many examples offunctions. The only kind of functions we will be concerned with in this class are functionsbetween sets of numbers. In fact, the real point of this section is to teach you a new wayto think about algebraic formulas.


Instead of thinking of a formula like 2x2 as something which stands for some unknownnumber, we will think of it as defining a function f which assigns, to every number x, thenew number f(x) = 2x2. It is like a machine that takes numbers as input and outputsnew numbers. So, in this case, f(x) eats 1 and turns it into 2(1)2 = 2, it eats −1 andturns it into 2(−1)2 = 2, it eats 2 and transforms it into 2(2)2 = 8. We no longer thinkof x as some fixed hidden number, but instead think of it as a slot to put numbers into.

Homework 3, # 1 Compute f(0), f(1), f(π), f(−2.2), and f(−10) for the followingfunctions.

(a) f(x) = 11−x

(b) f(x) =√x

(c) f(x) = x2−1x+10

For more practice, try problems 9-19 of Section 2.1.

There are two important definitions to remember for functions.

Definition 3.2. If f is a function from A to B, then A is called the domain of f . Inparticular, if f(x) is a function of numbers defined by an algebraic formula, then we alwaysassume the domain of f is the set of numbers for which the algebraic formula makes sense.

So, for example, the domain of f(x) = 1x−1 is the set of all numbers except 1. The do-

main of f(x) = x2 is all numbers because it always at least makes sense. Meanwhile thedomain of f(x) =

√x is the set of all non-negative numbers. To see why, let’s remember

what√x is in words:

√x is the non-negative number y such that y2 = x.

So√

0 = 0, since 02 = 0, and√

4 = 2, since 22 = 4. However, as it stands,√−1 is

undefined, because there is no real number y such that y2 = −1.

Definition 3.3. If f is a function from A to B, then B is said to be the range of f .In particular, if f(x) is a function of numbers defined by an algebraic formula, then wealways assume the range of f is the set of numbers y such that f(x) = y for some x inthe domain of f(x).

If we think of f(x) as something which shoots at numbers, then the range of f(x) isthe set of numbers which it actually hits. So, for example, if f(x) = x2, then when f(x)loads x = 2, it hits x = 4, and we conclude that 4 is in the range of f(x). Similarly,since f(3) = 32 = 9, when f(x) loads 3, it hits 9. So 9 is in the range of f(x). However,notice that no negative number is in the range of x2, because even if you plug in negativenumbers, only positive numbers pop out.

Homework 3,# 2 Find the domain and range of the following functions.

(a) f(x) = 2x+ 1(b) f(x) = −78x− 125(c) f(x) = 25− x2(d) f(x) = x−1

x


(e) f(x) =√x− 2

3.2. Graphs of Equations. Recall that Cartesian coordinates are a way to assign apair of numbers to every point in the plane. The way they work is simple. You draw ahorizontal line with notches in it corresponding to numbers, and call it the x-axis. Youthen draw a vertical line with numbered notches in it and call it the y-axis (it is alsocustomary to make the x-axis and y-axis meet at 0 on both). Do this on a separate pieceof paper.

Once these axes have been drawn, you have a way to associate a pair of numbers (a, b)to every point on the plane. The first number a is the vertical “shadow” cast by the pointon the x-axis, and the second number b is the horizontal “shadow” cast by the point onthe y-axis. You can also go in the reverse direction: starting with a pair of numbers (a, b)you can find the point on the plane which corresponds to it. Practice this with a coupleof points yourself.

This is probably one of the best ideas of all time, because it allows us to make aconnection between algebraic equations and shapes in space.

Definition 3.4. Suppose f(x, y) and g(x, y) are algebraic expressions in two unknowns,x and y. Then the graph of the equation

f(x, y) = g(x, y)

is the set of points (a, b) in the Cartesian plane which make the equation true when yousubstitute a for x and b for y.

So, for a simple example, the graph of the equation

(x+ y)2 = x2 + 2xy + y2

is the entire plane, because all pairs of numbers (a, b) make the equation true when youplug in a for x and b for y. Another simple example is the equation

x2 + y2 = −1

whose graph consists of no points at all! However, most graphs will be lines or curves ofsome kind. For example, the graph of the equation

(3.1) x2 + y2 = 4

is a circle of radius 2. To see why, notice that if (a, b) is a point on the graph, thena2 + b2 = 4 = 22. But if we set c = 2, then this becomes a2 + b2 = c2, the formula for thePythagorean theorem.

Homework 3, # 3: Draw the graphs of the following equations.

(a) x2 + y2 = 9(b) x+ y = 1(c) xy = 1(d) 2x− y = −1(e) x− 1

4y = 2

(f) 2x2 + y2 = 4


For more practice, try problems 11-26 of Section 1.3 (some of them are harder than youneed to know, but worth trying).

3.3. Graphs of Functions.

Definition 3.5. Suppose f(x) is a function. Then the graph of f(x) is just the graph ofthe equation

y = f(x).

In other words, the graph of f(x) is the set of points of the form (x, f(x)). It’s easyto use the “vertical line test” to determine whether or not a curve in the plane is thegraph of a function, and it’s also easy to figure out the domain and range of a functionby looking at its graph.

Homework 3, # 4

(a) Explain in your own words why a vertical line can only intersect the graph of afunction in one point at most.

(b) Suppose G is the graph of a function f(x). Explain why the domain of f(x) is the“shadow” of G on the x-axis, and the range of f(x) is the “shadow” of G on they-axis.

4. Linear Functions

Today we finish up Section 2.2, then move on to Section 2.4.

4.1. A bit more about the graphs of general functions. Last class period I endedby introducing graphs of functions, and had you think about three basic facts:

(1) A vertical line can only intersect the graph of a function in at most one point(Vertical Line Test).

(2) The domain of a function is the shadow that its graph casts on the x-axis.(3) The range of a function is the shadow that its graph casts on the y-axis.

Let’s put these ideas to use now.

Homework 4, # 1 I’m going to draw six graphs on the board. For each of them, dothe following:

(a) Use the vertical line test to determine whether it is a function.(b) If it is a function, find its domain.(c) If it is a function, find its range.

For more practice, try problems 11-18 and 31-40 of Chapter 2.2 of the book.

4.2. Linear functions and the slope-intercept form. Suppose A, B, and C areconstants. Then the graph of the equation

(4.1) Ax+By = C


will always be a straight line.

Homework 4, # 2 Verify this by plotting the graphs of the following equations.

(a) 2x− y = 1(b) 3x = 6(c) 8x+ 4y = 0

For more practice, try problems 1-14 in Section 2.4 of the book.

We can almost1 always rewrite Equation 4.1 so that it defines y as a function of x:

Ax+By = C

By = −Ax+ C

y = −ABx+

C

B

(4.2)

Those of you who have seen this before will recognize that this is just the “slope-intercept” equation of a line, which is normally written like this:

(4.3) y = mx+ b

Where m and b are constants. Notice that if b = 0, then this equation just becomesy = mx.

Homework 4, # 3 Draw the graph of the following functions.

(a) y = 2x(b) y = 1

2x

(c) y = −2x

Notice that these lines all pass through the point (0, 0), but they point in differentdirections. In general, the direction a line points in is determined by the constant m inEquation 4.3, which is called the slope.

Slope Rule: If a line has slope m, then for every unit you move right in the x-direction,you move m units vertically in the y direction.

If the slope m is written as a fraction cd, where d is a positive number, then we can say

that the line “rises”2 by c units for every d units you move right. In any event, for thefunctions in Problem 3 above, you could always use the following procedure to find thegraph:

(1) Draw a dot at (0, 0)

1The only time the algebra won’t work is if B = 0, but in that case Equation 4.1 becomes Ax = C,whose graph is just the vertical line of points (x, y) with x = C

A .2Although if c is a negative number it will actually fall by |c| units as you move right by d units.


(2) Move right 1 unit, and vertically m units, then draw another dot.(3) Draw the straight line through these two dots.

Now it is a general fact about functions that the graph of f(x) + b is what you get bymoving f(x) up by b units (or down if b is negative). So, to draw the graph of a linearfunction of the form y = mx + b, we only need to change the first step of the previousprocedure:

(1) Draw a dot at (0, b)(2) Move right 1 unit, and vertically m units, then draw another dot.(3) Draw the straight line through these two dots.

Use this procedure to solve the following homework problem.

Homework 4, #4

(a) y = 2x+ 1(b) y = −1

3x+ 3

(c) f(x) = 34x− 2

For more practice on problems like these, try problems 29-40 in Section 2.4 of the book.

Since it’s so easy to graph an equation in slope-intercept form, it usually saves a lot oftime to simplify a linear equation down to slope-intercept form first before graphing it.

Homework 4, #5 Graph these equations by putting them into slope-intercept form first(and if you can’t for some reason–improvise!).

(a) 2x− 13y = 2

3(b) 7x = −3 + y(c) 1

4y + 2x = 1− x

(d) 4x = −12


5. The point-slope form, finding the equation of the line.

In this part of the notes we discuss Sections 2.4 (a bit more) and Section 2.5 of thebook.

5.1. Going from line to equation, finding the slope. In the previous class periodwe saw how to change a linear equation into slope-intercept form.

(5.1) y = mx+ b

Where m tells us the slope and b is the y-intercept of the line. Once we know m and b,it is easy to find the line.

The slope-intercept form is also nice when you want to solve the reverse problem: start-ing with the graph of a line function, find an equation for it. All you need to be able todo is read off the line’s y-intercept and determine its slope. Recall how finding the slope


just means finding the “rise over run” of the graph.

HW 5, # 1: I am going to draw three graphs on the board. Find the slope-intercept formfor each of them.

Exercises 27 and 28 of Section 2.4 each have three lines drawn on them. For morepractice, find their slope-intercept form as well.

Suppose you know the coordinates (x0, y0) and (x1, y1) of two points on the graph of aline, where x1 > x0. Then,

• the “run” of the line between the two points is x1 − x0 and;• the “rise” of the line between the two points is y1 − y0.

Since the slope is just the “rise over run,” this leads to a nice formula for the slope.

Slope Formula: Suppose that (x0, y0) and (x1, y2) are both points on the line l. Thenthe slope m of l is given by:

(5.2) m =y1 − y0x1 − x0

Notice that it doesn’t matter which two points (x0, y0) and (x1, y1) that we choose onthe line l, the answer will always come out the same anyway.

Homework 5, # 2: I am going to draw three lines on the board, and label some pointsthat lie on them. Use these points and the slope formula of equation 5.2 to find their slope.

For more practice with the slope formula, try exercises 15-28 of Section 2.4 (most ofthese exercises don’t draw a graph for you, they just list two points on the line for you toplug into the slope formula).

5.2. The Point-Slope formula. In the future, when doing calculus you will often onlyknow the following two things about a line:

• The slope of the line; and• some arbitrary point on the line.

Using this information, you will then need to find an equation describing the line. Thefollowing simple formula will always do the trick:

Point-Slope formula: Suppose that a line l has slope m and passes through the point(a, b). Then the following is an equation of the line:

(5.3) y − b = m(x− a)

The parentheses in this formula are important! If we simplify the right side of Equation5.3, we get mx−ma, not mx− a!


Homework 5, # 3 Find the slope-intercept form of the following lines. Hint: Start byfinding the point-slope form, then use algebra to switch it to slope-intercept.

(a) l passes through (−1, 2), has slope m = 12

(b) l passes through (1, 0), has slope m = −4(c) l passes through (2,−2) and (4,−1)(d) l passes through (1, 1), has slope m = 3.14(e) l passes through (1, 2) and (0,−1)(f) l has x-intercept 3, and y-intercept 4.

Notice that on parts (c), (e), and (f), you first need to use the Slope Formula fromEquation 5.2. For more practice, try problems 1-28 of Section 2.5.

You might be wondering, how do we know that the point-slope formula works? Well,once we know the slope m of the line, and a single point (a, b) on the line, then the slopeformula given in Equation 5.2 tells us that, for any other point (x, y) on the line, we musthave:

(5.4) m =y − bx− a

Where here we have substituted (x, y) for (x1, y1), and (a, b) for (x0, y0) in the slopeformula. After we multiply both sides of this equation by (x− a), we get

(5.5) (x− a)m = y − b

And this is just the point-slope formula written in reverse order.

5.3. One more little thing about parallel and perpendicular lines. A diagram Idraw on the board (together with a little bit of deductive logic), tells us the following fact:

Perpendicular Line Slope Formula: Suppose l has slope m and that the line p isperpendicular to l. Then the slope of p is −1

m.

On the other hand, it’s pretty easy to see that parallel lines will have the same slope.These facts, together with the point-slope formula, allows us to solve the following kindof problem.Homework 5, # 4 Find equations for the following lines:

(a) l is parallel to y = 2x but passes through (4, 2).(b) l passes through the point (−2,−7) and is perpendicular to the line y = −5x+ 4.(c) l passes through (−2, 2) and is parallel to the line with equation 2x− 3y = 7.(d) l passes through (−2, 2) but is perpendicular to the line with equation 2x−3y = 7.



6. Solving small systems of linear equations

In this part of the notes we cover Chapter 3.1 of the book.

6.1. Solutions of Systems of Equations. A system of equations in 2 variables is justa collection of equations in which the same 2 variables occur (and we will almost alwayschoose these variables to be x and y). For example:

(6.1)−y = x2

1 = xy

A solution of a system of equations in 2 variables is a pair of numbers (a, b) which makeevery equation in the system come out true when you substitute a for x and b for y. Forthe system above, it is easy to check by plugging in that the pair (−1,−1) is a solution,while (1, 1) is not.

Homework 6, #1 Verify that (−1,−1) is a solution to the system above by actuallyplugging in x = −1, y = −1. Also, show that (2, 3), (1, 1), and (−1, 1) are not solutionsby plugging them in. Then plug these same pairs of numbers into the system

(6.2)x2 = 2− y2x = −y

Which pairs are solutions, and which are not?

Usually, there will only be a handful of solutions to a system of equations, if any.

6.2. Finding solutions: the substitution method. It is often difficult or even impos-sible to find the solutions to a system of equations, if it is too complicated. One way totry, however, is to use the substitution method.

The substitution method: Suppose you have a system of two equations in two un-knowns, x and y. Then the following steps will lead to a solution:

(1) Use algebra to reduce one equation to the form y = f(x),.(2) Plug in f(x) for y in the other equation, then solve for x.(3) Plug each value of x back into y = f(x) to find corresponding y values.

Rather than explain with more words, it is easier to explain the process by example:

Example 6.1. Find all solutions of the original example system

−y = x2

1 = xy

Solution: The first equation can be rewritten as

y = −x2.Plugging in −x2 for y in the second equation gives us

1 = −x3.


Multiplying both sides by −1 then turns it into

−1 = x3

and by taking the cube root of both sides we find x = −1. Finally, plugging x = −1 backinto the first equation gives

y = −(−1)2 = −1.

Therefore there is only one solution, and it is given by the pair (−1, 1).

Homework 6, #2 Use the substitution method to find all solutions to the system

x2 = 2− y2x = −y

For more complicated systems of equations the substitution method won’t work, be-cause you won’t even be able to do step (1),or maybe step (2).

6.3. Using graphs to understand systems of equations. As we discussed in ear-lier sections, any equation in two unknowns x and y has a graph of points in the plane.This graph consists of all points (a, b) on the plane which make the equation come out true.

If we have a system of two equations E1 and E2 in two unknowns, then we will havetwo graphs, call them G1 and G2. If the point (a, b) lies in G1 and G2, then the pair ofnumbers x = a, y = b solves E1 and E2. And on the other hand, if the pair of numbersx = a, y = b solves both E1 and E2, then the point (a, b) lies on both graphs G1 and G2.Therefore we have the following basic fact:

FACT: If G1 and G2 are the graphs corresponding to a system of two equations in twounknowns, then the solutions of this system correspond to intersection points of G1 and G2.

Homework 6, # 3 Draw the graphs corresponding to both systems of equations oc-curring earlier in the notes. Verify that the intersections do indeed correspond to thesolutions we found earlier.

6.4. Linear systems of equations. For more complicated systems of equations the sub-stitution method won’t work because you just can’t do step (1), or maybe you get stuckon step (2). Using graphs to find solutions is also very hard. However, for the time beingyou will only need to be concerned with a very special kind of systems: Linear Systems.For linear systems, it is easy to find the graphs, and the substitution method always works.

Definition 6.2. A system of equations in two variables is said to be linear if everyequation in the system can be algebraically reduced to the form:

(6.3) Ax+By = C

Where, A, B, and C are constants.


Neither of the two systems we have been working with so far have been linear. However,all of the systems in the following homework problem are.

Homework 6, #4 Find all solutions (if any) to the following systems of equations usingthe substitution method.

(a)x+ y = 6

y = 2x

(b)x− 3y = 3

3x+ 5y = −19

(c)y = 2

5x− 2

2x− 5y = 10

For more practice, try problems 25-42 in Chapter 3.1 of the book.

Homework 6, #5 Use the graphing method to find all solutions (if any) to the followingsystems of equations.

(a)x+ y = 4x− y = 2

(b)x− y = 2

y = 1

(c)2x− 3y = 64x− 6y = 24

Homework 6, #6 In your own words, explain why there are only three possibilities fora system of linear equations in two unknowns:

(1) There are no solutions.(2) There is exactly one solution.(3) There are infinitely many solutions.

Hint: Use the big fact from Section 6.3 of these notes.

In case (1), the system is called inconsistent, and in the third case the system is saidto have a dependency, although the reasons for this latter terminology will not be clearto you for awhile.

7. Three variables.

In this part of the notes we finish up Section 3.1 and cover Section 3.3 as well.

7.1. The addition method for two variables. Besides the substitution and graphingmethods, there is yet another way to solve systems of linear equations. This method isusually faster than the other two methods, and it extends easily to larger systems of linearequations in more variables.


Suppose we have a system of two linear equations in two variables as in the last section:

(7.1)A1x+B1y = C1

A2x+B2y = C2

Recall that a pair of numbers (a, b) is a solution to this system if both equations comeout true when a is substituted for x and b for y. In addition to the usual rules for sim-plifying equations that you already know, there is one new move that we can perform onsystems of equations which does not change its set of solutions.

Addition Rule: In the system of equations 7.1 above, the first equation tells us thatA1x+B1y is interchangeable with C1 (that’s what it means for them to be equal!). There-fore, we can add any multiple n of A1x + B1y to the left side of the bottom equation,while adding the same multiple n of C1 to its right side without changing the solutionsto the system:

(7.2)A1x+B1y = C1

n(A1x+B1y) + A2x+B2y = C2 + nC1

And this simplifies to:

(7.3)A1x + B1y = C1

(nA1 + A2)x + (nB1 +B2)y = nC1 + C2

In other words, you can always add a multiple of the top row to the bottom row andstill get the same solutions. Of course, we can also add or subtract the bottom row fromthe top as well. This new rule, when used with other basic algebraic rules, is very powerful.

Example 7.1. Consider the system

(7.4)x+ y = 4x− y = 2

from your previous homework. We use the addition rule to add the top equation to thebottom one, and get:

(7.5)x+ y = 4

2x = 6

If we divide both sides of the bottom equation by 2 (which also does not change theset of solutions) we get:

(7.6)x+ y = 4

x = 3

Finally, we can subtract the bottom equation from the top, and get:

(7.7)y = 1x = 3


Homework 7, #1 Solve the same systems of equations as in Homework 6, #4 and #5,but use the addition method (it’s probably a good idea to start with the problems of #5first, because they are easier).

For more practice with the addition method in two variables, try exercises 43-58 ofChapter 3.1 in the book.

7.2. Systems of Linear Equations in 3 variables. Now let’s consider systems of threelinear equations in three unknowns x, y, and z, which have the following form:

(7.8)A1x+B1y + C1z = D1

A2x+B2y + C2z = D2

A3x+B3y + C3z = D3

Here, all the capital letters are assumed to be constants. A solution to this system is atriple of numbers (a, b, c) such that all three equations come out true when you substitutea for x, b for y, and c for z.

All of the reasoning is the same for these systems as it is for the systems in two vari-ables. In particular, you can still use the substitution method and the addition method(and there is even something like a “graphing method”) to find solutions to the equations.However, the substitution method is now usually much harder than the addition method.

Example 7.2. Let us solve the following system of equations:

(7.9)x+ y + 2z = 11x+ y + 3z = 14x+ 2y − z = 5

Just for the sake of teaching you a lesson, I’m going to solve this system in class usingthe substitution method. However, I’m not going to torment myself by writing it all outhere, especially since you should probably never use the substitution method on a systemthis big. Instead, I will use the addition method to quickly solve this equation. Noticefirst that if we subtract the top equation from the middle equation we get:

(7.10)x+ y + 2z = 11

z = 3x+ 2y − z = 5

If we add the middle equation to the bottom equation, and then also subtract 2 timesthe middle equation from the top equation, we get:

(7.11)x+ y = 5

z = 3x+ 2y = 8

Now if we subtract the top equation from the bottom, we get:


(7.12)x+ y = 5

z = 3y = 3

Finally, if we subtract the bottom from the top, we end with:

(7.13)x = 2z = 3y = 3

This tells us that the original equation had only one solution, the triple (2, 3, 3). Wecan check that this is right by actually plugging it into the original system!

Homework 7, #2 Solve the following systems of equations:

(a)

2x+ y − 2z = −1

3x− 3y − z = 5

x− 2y + 3z = 6

(b)

2x+ y = 2

x+ y − z = 4

3x+ 2y + z = 0

(c)

2x+ y + 2z = 1

3x− y + z = 2

x− 2y − z = 0

(d)

7z − 3 = 2(x− 3y)

5y + 3z − 7 = 4x

4 + 5z = 3(2x− y)

For more practice do problems 5-22 of Chapter 3.3 in the text.

7.3. The geometric significance of linear equations in 3 variables. we can useCartesian coordinates to associate a pair of numbers with every point on the plane, wecan use Cartesian coordinates to associate a triple of numbers to every point in space.The graph of an equation in three variables

f(x, y, z) = g(x, y, z)

is then just the set of points (a, b, c) in space which make the equation come out truewhen a = x, b = y, and c = z. When an equation has the form

Ax+By + Cz = D

its graph is guaranteed to be a plane. The set of solutions to a system of three suchequations as in 7.8 above is then the set of all points in space which lie on all three planes.It follows that there are only three possibilities for the number of solutions to a systemof three equations in three unknowns.

(1) There is exactly one solution. (This is the most common).


(2) There are infinitely many solutions.(3) There are no solutions.

8. Applications of Systems of Equations

In this part of the notes we cover Section 3.2 and a little more of 3.3.

8.1. The Fruit Stand. The ability to solve systems of linear equations is often usefulwhen you are trying to isolate the individual effects of two or more quantities whichalways work together. What I mean is best shown by the following contrived, but easilyunderstood, “fruit stand” word problem.

Example 8.1. John goes to a fruit stand and pays a total of $7 for two apples and threeoranges. Jane goes to the same fruit stand and pays a total of $6 for one apple and fouroranges. What was the price of apples, and what was the price of oranges?

A good first step with word problems is to identify important unknown quantities. Inthis problem, the important unknown quantities are the price of apples and the price oforanges. Let x denote the price of apples, and let y stand for the price of oranges, bothin dollars.

The next step is to convert the facts we have into equations. In this case, the firstsentence tells us that

2x+ 3y = 7

The second sentence tells us that

x+ 4y = 6.

The price of apples x and the price of oranges y will have to make both of these equationscome out true. Thus, we have now finally converted our word problem into a mathproblem: we must solve the system

(8.1)

{2x+ 3y = 7

x+ 4y = 6

If we subtract twice the bottom row from the first, we get the new system

(8.2)

{−5y = −5

x+ 4y = 6

From the top equation we see y = 1, and plugging this into the bottom equation tellsus x = 2. It follows that apples must have cost $2, and oranges $1.

Homework 8, #1 John pays $12 for three bananas, two mangos, and two boxes of straw-berries. Jane pays $10 for five bananas and two boxes of strawberries. Javier pays $20for five mangos and four boxes of strawberries. What is the price of bananas?

There are many problems which are just “fruit stand” problems in disguise. Here’s onefor you to think about over the weekend:


Homework 8, #2 Rex the dog likes two brands of dog food: bacon-flavored and steak-flavored. Bacon-flavored costs $10 a bag, and steak-flavored costs $15 a bag. He endedup eating 10 total bags of dog food last month, on which you spent a total of $120. Howmany of the bags that he ate were bacon-flavored?

Finally, I would like you to know how to do this kind of problem:Homework 8, #3 You invested $7000 in two accounts paying 6% and 8% annual interest.If the total interest you earned for the year was $520, how much was invested at each rate?

For full homework credit on this I need to see some work, not just an answer. For morepractice, try problems 12-18, and 25-28 in Section 3.2 of the book.

8.2. Finding parameters of geometric shapes. Any parabola with a vertical axis ofsymmetry will be the graph of a function of the form f(x) = ax2 + bx+ c. Suppose nowthat someone just draws the graph of such a parabola, and gives you the coordinates ofthree points on it. Can you use these to determine the full of equation of the parabola?

The answer is yes, using systems of linear equations.

Example 8.2. Suppose a parabola with vertical axis of symmetry passes through the points(−1, 2), (0, 4) and (1, 1). Find the equation of this parabola.

We know that, whatever the equation is, it has the form y = ax2 + bx + c. The un-knowns for us in this situation are the numbers a, b, and c. Once we find them, we findthe equation.

Knowing that the parabola passes through (−1, 2) tells us that the equation

2 = a(−1)2 + b(−1) + c

is true, which is the same as2 = a− b+ c.

Similarly, the fact that (0, 4) lies on the parabola tells us that

4 = a · 02 + b · 0 + c.

So in fact c = 4. Finally, since (1, 1) lies on the graph, we see that

1 = a · 12 + b · 1 + c.

All told then, we see that we want a solution of the system

(8.3)

a− b+ c = 2

c = 4

a+ b+ c = 1

If we subtract the second equation from the top and bottom, we get

(8.4)

a− b = −2

c = 4

a+ b = −3


If we add the bottom to the top we get

(8.5)

2a = −5

c = 4

a+ b = −3

From the first equation we see that a = −52

, plugging this into the bottom we see b = 12,

and from the middle equation we immediately see that c = 4. Therefore the equation ofthe parabola that we are after is

y =−5

2x2 +

1

2x+ 4.

Homework 8, #4 Find the equation y = ax2 + bx + c of the parabolas which passthrough the following triples of points. If you cannot find them, explain why.

(a) (−1, 4), (0, 1), (2, 9).(b) (−2,−1), (1, 1), (2,−4).(c) (0, 2), (0, 3), (1, 4)

For more practice, try problems 31 and 32 in Section 3.3 of the book.

Similarly, any equation of the form

ax2 + by2 = 1

gives us the equation of an ellipse centered at zero with vertical and horizontal axes ofsymmetry. It makes intuitive sense that any such ellipse will be determined by two points,and using the same method as above (except easier) we can find its equation.

Homework 8, #5 Find the equation of the ellipse ax2 + by2 = 1 which passes throughthe following pairs of points. If you cannot find such an ellipse, explain why. If there aremany, explain why.

(a) (4, 0), (0, 1)(b) (1,−1), (2, 0)(c) (1,−1), (1, 1)(d) (1,−1), (1, 1.1)

9. Inequalities

This section of the notes covers 4.1-4.3 of the text.

9.1. The meaning of an inequality. We say that a < b if there is some positive numberc such that a+ c = b. We say that a ≤ b if a < b or a = b. Finally, a > b and a ≥ b meanthe same thing as b < a and b ≤ a, respectively.

Example 9.1. The following inequalities are true:

• 2 < 4


• π ≤ 4• −2 > −4

Definition 9.2. A number a is a solution of the inequality

f(x) < g(x)

if the inequality comes out true when you substitute a for x. Similar definitions apply forthe symbols >, ≤, and ≥.

When you solve an equation of the form f(x) = g(x), there are usually only a handful ofsolutions at most. However there are usually an infinite number of solutions to inequalities.

Example 9.3. The inequality 2x < x2 is true whenever x > 2 or x < 0.

You might be wondering how I figured that out. The basic idea is to use algebra tosimplify the inequality like you would with an equation. For example, if a < b, then forany number c, a+ c < b+ c. Therefore, given any inequality

f(x) < g(x)

we can add a function h(x) to both sides and get a new inequality

f(x) + h(x) < g(x) + h(x)

which has the same solutions. For example, using this fact we can simplify the inequality

2x < x2

like so

(−2x) + 2x < x2 − 2x

0 < x2 − 2x

0 < x(x− 2)

(9.1)

And in this last simple form it is easy to see that the inequality is true when x < 0 or2 < x. Notice, however, that if 0 < x < 2, then x is positive while (x− 2) ≤ 0 is negative,which forces x(x− 2) to be negative.

Homework 9.1 Solve the following inequalities:

(a) 2x− 11 < 3 + x(b) 4(x+ 2)− 3 ≥ 2x+ 8(c) x−2

3> 2x+1

4

(d) 1−x−3 ≤ 2x+ 1

Notice that you can partially verify whether your answer is correct by plugging in a fewvalues of x and seeing if they work. For more practice try problems 1-38 of Section 4.1.


9.2. A warning about multiplication and division. It is tempting to try to solve aninequality like

2x < x2

by dividing both sides of the equation by x, which turns it into

2 < x.

But notice that this simplification actually throws out all the negative numbers that wefound to be solutions earlier! This happened because we divided both sides of the in-equality by the variable x.

WARNING: You are not allowed to multiply or divide both sides of an equality by x!

If a < b then it is only true that ac < bc if c is a positive number. If c is negative, thena < b implies ac > bc. For example, 1 < 2, and if we multiply both sides of this inequalityby 2 we get 2 < 4, which is fine. However, if we multiply both sides of this inequality byc = −2 we have to reverse the inequality to get −2 > −4.

Special Rules for Multiplication and Division: When solving an inequality f(x) <g(x), we can perform the following moves:

(1) Changing it into Cf(x) < Cg(x) or f(x)C

< g(x)C

if C is a positive constant.

(2) Changing it into Cf(x) > Cg(x) or f(x)C

> g(x)C

when C is a negative constant.

We cannot multiply or divide both sides of an inequality by a variable x because itmight stand for a positive C or a negative C, and the rules are different for positive andnegative C.

Homework 9.2: Solve the following inequalities.

(a) −3x < −8(b) −2x+1

−2 > 1

(c) 2x−33≥ 1− 2

3x

(d) 3x2 > 2x(e) 2x−1

x> 1

For more practice you can try the same problems that I listed after HW 9.2 of thesenotes. They will be easier than some of the problems above, but they are as hard as youneed to know for the quizzes.

9.3. How to deal with the words “and” and “or” with inequalities. Solving a“compound inequality” just means solving a couple of inequalities at the same time. Forexample:

Example 9.4. Find the set of all x such that 2x− 1 > 3 AND x− 1 ≤ 8.

Solution: The inequality 2x−1 > 3 reduces to x > 2, and the inequality x−1 ≤ 8 reducesto x ≤ 9. Therefore, the set of all x which make both true is the set 2 < x ≤ 9, or (2, 9],


in interval notation.

When solving compound inequalities, it is useful to draw a number line to keep trackof things. When solving a compound inequality involving the word “AND” as above, theset of solutions will be the overlap of the sets of each equation you solve.

If you are asked to find all solutions to g(x) < f(x) < h(x), this is the same as the setof x which solve g(x) < f(x) AND f(x) < h(x).

Homework 9.3: Find the set of all x such that:

(a) 1−4x2

< 1 and 3x− 5 ≤ 6(b) −6 < x− 4 ≤ 1(c) 2 ≤ x−4

3≤ 2

(d) −1 < 1− x ≤ 7.

Express your answers in interval notation. For more practice try problems 15-32 ofSection 4.2.

Here’s a somewhat different kind of compound inequality.

Example 9.5. Find the set of all x which solve 2x− 3 > 2− x OR 1−xx≤ 2.

Solution: The first inequality reduces to x > 53. The second inequality reduces to 1

x≤ 3,

which is solved by all x ≤ 13

such that x 6= 0. In interval notation, then, our solution is

(−∞, 0) ∪ (0, 13) ∪ (5

3,∞).

Again, it is useful to use number lines to keep track of everything. The difference isthat, with compound inequalities which use the word OR, the set of solutions will consistof all the solutions that occur in either inequality, not just their overlap.

Homework 9.4 Find the set of all x such that:

(a) x < 3 or x ≤ −1(b) 3x+ 2 ≤ 5 or 5x− 7 ≥ 8(c) x2 ≥ 9 or x2 < 1(d) 1

1−x > 0 or x ≥ 1.

Express your answers in interval notation. For more practice try problems 39-58.

10. The absolute value

This section of the notes covers Section 4.3 of the notes.

10.1. Solving equations involving the absolute value. The absolute value function|x| is defined to be x when x > 0 and −x when x < 0. Therefore, |x| = 3 when x = 3 ORwhen x = −3. Similarly, more complicated equations like

|2x+ 1| = 3

might be true when2x+ 1 = 3


OR−(2x+ 1) = 3,

but you need to plug in at the end to make sure the solutions you find really do work.Here we find the two solutions x = 1 and x = −2 both work in the original.

Example 10.1. Find the solutions of 2x|1− 2x|+ x = 2x− 3x2.

Solution: It usually saves time to first reduce the equation to the form |f(x)| = g(x) first.

2x|1− 2x|+ x = 2x− 3x2

2x|1− 2x| = x− 3x2

|1− 2x| = 1

2− 3

2x

(10.1)

Now that it is in this form, we break it up into two cases. When |1− 2x| = 1− 2x, theequation becomes

1− 2x =1

2− 3

2x

1

2=

1

2x

1 = x

(10.2)

However, notice that the solution x = 1 does not actually work in the original equation!This is because |1− 2x| = 1− 2x is not true when x = 1, like we assumed at first. So infact the equation has no solution when |1− 2x| = 1− 2x.

The second case to consider is |1− 2x| = −(1− 2x) = 2x− 1. In this case our equationbecomes

2x− 1 =1

2− 3

2x

7

2x =

3

2

x =3

7

(10.3)

You can now verify that x = 37

is a solution. By the way, x = 0 is also a solution, onewhich we forgot the moment we divided both sides of the original equation by 2x. Thinkabout that.

Homework 10.1: Solve for x.

(a) |x| = −1(b) |x− 1| = 4(c) |3x− 2| = 5− x(d) 2|x+ 1| − 3 = 4− 2x(e) 3|2x+ 1| = |3− 2x|


Try 1-36 from Section 4.3 for more practice.

10.2. Solving inequalities involving the absolute value function. It is easy to seethat |x| < C is true when −C < x < C, and |x| > C when x > C or x < −C (assumingC is a positive number). The same things are true for algebraic expressions |f(x)|. Thisis the main fact we use when solving inequalities involving |f(x)|.

Example 10.2. Find the solutions of |2x+63| < 2.

Solution: The set of x which solve this is the same as the set of x which solve the compoundinequality

(10.4)

−2 < 2x+63

< 2−6 < 2x+ 6 < 6−12 < 2x < 0−6 < x < 0

Thus, our set of solutions is the interval (−6, 0).

Homework 10.2: Find all solutions of the following inequalities.

(a) |x− 2| > 1(b) |3x− 8| ≥ 5(c) |3x−3

4| < 6

(d) |x− 2|+ 4 ≤ 5(e) −2|5− x| < −6


11. Polynomials, Part I

This lecture corresponds to Sections 5.1, 5.2, and 5.3 of the book.

11.1. Some terminology. A monomial is an expression of the form Kxnymzp . . ., whereK is a constant, n,m, p are non-negative whole numbers, and x, y, and z are variables(we allow more variables than these to occur, and we also allow fewer). Technically wealso count a single number, like 5, as a monomial.

Example 11.1. The following expressions are monomials:

• x• −1.5y3z2

• π2xy3z2wu25

• 3y2

• 2π

The following expressions are not monomials:

• x1.5• 2xy−2

• x+ 5


• 3√x

• 2xy

A polynomial is a sum of monomials.

Example 11.2. Here are some polynomials:

• Any monomial• x2 − 2x+ 1• 3xy2 − πx4y25z − 2• x+ 5

The monomials which make up a polynomial are called the terms of the polynomial.The coefficient of a term Kxnymzp . . . is the constant K. Two terms are said to be liketerms if they are the same, except for their coefficients. So 2x4y3 and −7x4y3 are liketerms, but 2x4y3 and 2x3y3 are not like terms. A polynomial is in standard form if nolike terms appear inside of it.

Example 11.3. The following polynomials are in standard form:

• 2x2 − 3xy + 2xy2 − y2 + 2• 7x3 − 2x+ 1• (2 + π)x2 − y

These ones are not:

• 3x2 + x2 − x• πxy + x3 − 2x− 1.5yx

11.2. Adding and Subtracting Polynomials. Recall the distributive law of numbers:

ba+ ca = (b+ c)a.

This rule tells us that if you have two like terms, you can combine them into a single termby adding their coefficients.

Example 11.4. • 2x3 + 3x3 = 5x3

• 7xy2 − xy2 = 6xy2

• 3x2z − 7x2z = −4x2z

If you add or subtract polynomials in standard form, what you get out will be a poly-nomial, but not necessarily one in standard form. The put the result in standard formyou need to add like terms using the rule above. When I ask you to add or subtract twopolynomials, that always means that I want the result in standard form.

Homework 11.1: Add or subtract the following polynomials.

(a) (3x2 − 7x+ 1) + (4x3 − 2x+ 5)(b) (4xy2 − 7x+ 2y2) + (2x2y + 5xy + 2y − 5y3)(c) (3x− 7xy + 5)− (35x2 − xy + πx− 1 + y2).(d) (5x4y2 + 6xyw − 3)− (6xy − y2x4 + 3)

For more practice on this try problems 29-54 of Section 5.1. Remember to distributeyour minus signs! That’s pretty much the only trick to this whole thing.


11.3. Multiplying polynomials. It follows from the basic exponent laws (as well as thecommutative and associative properties of multiplication), that you multiply monomialsaccording to the following equation:

(Kxnymzp)(Cxjykzl) = KCxn+jym+kzp+l

In other words, you multiply the coefficients and add the exponents.

Example 11.5. Multiplying monomials:

• (3x2y)(−2x3y4) = −6x5y5

• (12xyz3)(2xy2) = x2y3z3

• (−3x2yz7)(πzx)(2y2z5x) = −6πx4y3z13

To multiply polynomials, you multiply each term from one by each term of the other,and add the results up. Often, this will leave you with a polynomial that is not in stan-dard form, so you also have to add all like terms at the end.

Homework 11.2: Multiply the following polynomials out into standard form.

(a) 3x2y(z − 7x2 + 5x− 13)

(b) (x2 − xy)(2x− 3y2)(c) (−2x+ 5x2)(1− x+ 2x2)(d) (x− 2y − xy)(5x+ y + 1)(e) (x2 − 2x+ 3)(x− 1)(2x2 − 3)(f) (2x+ 1)2(3x− 1)

For additional practice, try problems 1-95 of Section 5.2 in the text. Multiplyingpolynomials is a core skill, you should be completely fluent with it.

11.4. Factoring Polynomials. In the last exercise we took two or more polynomialsp(x) and q(x) in standard form and multiplied them out to get a new polynomial f(x) instandard form. It is also useful to be able to go the opposite direction: starting with asingle polynomial f(x) in standard form, rewrite it in the form (p(x)) · (q(x)), where p(x)and q(x) are polynomials. This process is called factoring f(x), and when it is carriedout, p(x) and q(x) are called factors of f(x).

Example 11.6. Here are some factorizations:

• x2 + 2x+ 1 = (x+ 1)(x+ 1) = (x+ 1)2

• x2 − 9 = (x+ 3)(x− 3)• x3 − x2 + x− 1 = (x− 1)(x+ 1)2

• x3 + 3x2 + 4x+ 2 = (x+ 1)(x2 + 2x+ 2)• x4 − 6x3 + 16x2 − 22x+ 15 = (x2 − 4x+ 5)(x2 − 2x+ 3)

As these examples show, multiplying polynomials out is easy, factoring them is hard. Itis easy to check that if you multiply out (x+1)(x2+2x+2), you will get out x3+3x2+4x+2,but if I just gave you the polynomial x3 + 3x2 + 4x + 2 and asked you to factor it, myguess is that few if any of you could do it (at least not until you learn more). And even Icouldn’t factor the polynomial x4− 6x3 + 16x2− 22x+ 15 out if someone gave it to me. Icame up with the example by first thinking up the factorization, then multiplying it out.

Because it is so hard, the rest of Chapter 5 in the textbook is devoted just to factoringspecial forms of polynomials. The easiest form of factoring is factoring out a commonterm.


Example 11.7. Every term in the polynomial p(x) = 9x3 − 3x2 + 6x has the monomial3x as a factor. Thus we can use the distributive property ab+ ac = a(b+ c) to “pull out”the 3x to get p(x) = 3x(3x2 − x+ 2)

Homework 11.3: Factor out the largest common factor of the following polynomials:

(a) 7x2 + 3x3

(b) 2x− 4x2 + 8x4

(c) −x4 + 2x3 + 7x2

(d) 3x2 − 9x4 + 2x

For more practice, try problems 1-34 of Section 5.3.A more subtle factorization technique (and one which only works occasionally in real

life) is factorization by grouping. This is best illustrated by example.

Example 11.8. To factorx3 − 2x2 + 5x− 10,

we first “group” x3 − 2x2 and 5x − 10 together, and pull out the largest common factorfrom each group to obtain

x2(x− 2) + 5(x− 2).

But lo, a miracle has occured: the polynomial (x− 2) is in both terms now. This allowsus to use the distributive law ba + ca = (b + c)a (setting a = (x − 2)) to get out thefactorization

(x2 + 5)(x− 2).

The reason factorization by grouping usually doesn’t work is that you won’t get a com-mon thing like (x− 2) occuring in both terms after factoring. But when all else fails, it isworth a try at least. And it’s good practice for understanding the logic of factorization.Here are some examples where factoring by grouping does work:

Homework 11.4: Factor these by grouping.

(a) x2 + 3x+ 5x+ 15(b) xy − 5x+ 9y − 45(c) 3x2 − 6xy + 5xy − 10y2

(d) 2x3 − 10 + 4x2 − 5x

For more practice, try problems 45-68 of Chapter 5.3.

12. Polynomials, Part II

In this part of the notes we cover Sections 5.4-5.6 of the book, although these noteswill go a bit beyond what is covered there (I’m going to bring up the quadratic formula,which the book doesn’t cover until Section 8.2–you will understand why soon).

12.1. A primitive method for factoring various kinds of trinomials. Trinomial isa fancy word we use for a polynomial which has 3 terms, because why not? The first kindof trinomial you need to learn how to factor has the following form:

(12.1) p(x) = x2 + ax+ b

If such a polynomial can be factored at all, its factorization will have this form:


(12.2) p(x) = (x+ c)(x+ d)

If we multiply this last equation out, it becomes x2 + (c+ d)x+ cd. Therefore, makingthe coefficients match with equation 12.1, we get the following two equations out:

(1) c+ d = a(2) cd = b

Therefore, factoring a trinomial of the form 12.1 boils down to solving this system ofequations, which we do by guessing and checking. Let’s look at a real trinomial to seehow this works.

Example 12.1. Let’s factor the polynomial

x2 − 2x− 8.

We know that our answer will have to have the form (x − c)(x − d) for some pair ofnumbers c and d. Not only that, we know from the equations above that c+ d = −2 andcd = −8. Assuming that both c and d are whole numbers, we write out a list of “guess”numbers for c and d which make cd = −8.

• c = −1, d = 8• c = 1, d = −8• c = −2, d = 4• c = 2, d = −4

Now we check each pair to see if it solves the equation c + d = −2. The winning pairis c = 2, d = −4. So we obtain the factorization

x2 − 2x− 8 = (x+ 2)(x− 4).

Homework 12.1: Factor the following polynomials:

(a) x2 − x− 6(b) x2 + 7x+ 10(c) x2 − 6x− 16(d) x2 − 6xy − 16y2

(e) x2 − 8xy + 15y2

Try problems 1-30 of Section 5.4 for more practice.

Let’s up the complication level a little bit and try factoring polynomials of the form

(12.3) p(x) = ax2 + bx+ c

Because we are now allowing the first coefficient to be something other than 1, thefactorizations we find might be crazier. They might have the form:

(12.4) p(x) = (k1x+ d1)(k2x+ d2)

Multiplying this out gives us k1k2x2 + (k1d2 + d1k2)x+ d1d2. Matching coefficients up

with our original polynomial in 12.3 now gives us three equations:

(1) k1k2 = a(2) d1d2 = c


(3) k1d2 + d1k2 = b

So now we can use basically the same guess and check method as we did in the simplercase, except with more junk to look at.

Example 12.2. Let’s factor3x2 + 3x− 18.

Our solution will have the form (k1x + d1)(k2x + d2), where k1k2 = 3, d1d2 = −18, andk1d2 + d1k2 = 3. The first equation gives us the following list of whole number candidatesfor k1 and k2.

• k1 = 1, k2 = 3• k1 = 3, k2 = 1

From the equation d1d2 = −18 we get the following candidates for d1 and d2:

• d1 = ±1, d2 = ∓18• d1 = ±2, d2 = ∓9• d1 = ±3, d2 = ∓6

Testing all possible pairs of pairs in the third equation k1d2 + d1k2 = 3, we end up withthe winning set k1 = 1, k2 = 3, d1 = 3, and d2 = −6. Thus we obtain the factorization

3x2 + 3x− 18 = (x+ 3)(3x− 6).

Homework 12.2: Factor the following polynomials.

(a) 3x2 + 8x+ 5(b) 6x2 + 19x+ 15(c) 6x2 − 7xy − 5y2

Try problems 45-68 of section 5.4 for more practice.

12.2. An automatic way to factor any trinomial that can be factored. If p(x) isa polynomial and x0 is a number such that p(x0) = 0, then x0 is called a root of p(x).

Proposition 12.3. If x0 is a root of p(x), then p(x) = (x− x0)q(x) for some polynomialq(x).

It follows that if the trinomial ax2 + bx+ c has roots x0 and x1, then it can be factoredas a(x − x0)(x − x1). On the other hand, if it has no roots, it cannot be factored at all.Luckily, we can always find the roots of a polynomial (if they exist) using the quadraticformula.

Proposition 12.4. The roots x0 of the polynomial ax2 + bx+ c are given by the formula

x0 =−b±

√b2 − 4ac

2a.

Example 12.5. Using the quadratic formula of Proposition 12.4, we see that the rootsof 5x2 − 3x− 2 are

x0 =−(−3)±

√(−3)2 − 4(5)(−2)

2(5)=

3±√

9 + 40

10=

3

10± 7

10= 1,−2

5

. Now, using Propositin 12.4, we get the factorization

5x2 − 3x− 2 = 5(x− 1)(x+2

5).


If you really hate fractions, you can also rewrite this last expression to obtain the factor-ization

5x2 − 3x− 2 = (x− 1)(5x+ 2).

Homework 12.3: Use the quadratic formula to find the roots of the following polyno-mials (if there are any), then factor them if possible.

(a) x2 − 3x+ 2(b) y2 − 3y − 40(c) 8x2 − 18x+ 9(d) 4x2 − 27x+ 18(e) x2 + 2x+ 3

The advantage of using the quadratic formula to factor a trinomial is that it worksautomatically (just plug and chug, no need for creative thought) and it works everytime, even when the polynomial you are factoring does not have integer solutions. Thedisadvantage is that the formula is complicated.

12.3. Special forms. Observe the following equations:

(1) A2 −B2 = (A+B)(A−B)(2) A3 +B3 = (A+B)(A2 − AB +B2)(3) A3 −B3 = (A−B)(A2 + AB −B2)

These formulas allow you to quickly factor differences of squares and sums or differ-ences of cubes. The cubes are especially important to remember, because none of themethods we’ve discussed so far can be used to factor them in general.

Homework 12.4: Factor these.

(a) 36x2 − 49y2

(b) x3 + 64(c) 125x6 − 64y3

Try 75-94 of Section 5.5 for more practice.

12.4. Three ways of factoring even bigger polynomials. Using what we’ve learnedso far, we can factor even bigger polynomials. The first way is to start by factoring outa common factor, then factor what’s left. Factoring out a common factor is one of those“gimme” things that you always do when you have the chance.

Example 12.6. Let’s factor 4x3−6x2 +2x. We start by factoring out the common factor2x to get 2x(2x2 − 3x + 1). We now factor 2x2 − 3x + 1. The quadratic formula tells usthat the roots of 2x2 − 3x+ 1 are

x =3±√

9− 8

4= 1,

1

2.

Therefore 2x2 − 3x+ 1 = 2(x− 12)(x− 1) = (2x− 1)(x− 1). We conclude

4x3 − 6x2 + 2x = 2x(2x− 1)(x− 1).

If that doesn’t work, you can try factoring by grouping first.


Example 12.7. Let’s factor x3− yx2 + 2x2− 2xy+ x− y. We’ll break it into two groupsof terms, the group of terms with y occuring in them, and all the rest, like so:

x3 + 2x2 + x− yx2 − 2xy − y = x(x2 + 2x+ 1)− y(x2 + 2x+ 1) = (x− y)(x2 + 2x+ 1).

Finally, it is quick to use the “primitive” method of factoring to show that x2+2x+1 =(x+ 1)2. Thus:

x3 − yx2 + 2x2 − 2xy + x− y = (x− y)(x+ 1)2

A third easy way is to make a substitution that reduces the polynomial to one that canbe factored.

Example 12.8. Let’s factor x4 + 2x2− 15. First, notice that it can be written as (x2)2 +2(x2) − 15, so setting y = x2 it becomes y2 + 2y − 15. This can be factored using thequadratic formula, its roots are:

y =−2±

√4− 4(−15)

2=−2± 2

√16

2= 3,−5

Therefore we get the factorization (y − 3)(y + 5). Substituting y = x2 back in, we get(x2 − 3)(x2 + 5). x2 − 3 is a difference of squares, and so can be factored using a formulax2 − 3 = (x+

√3)(x−

√3). x2 + 5 is prime. Therefore we obtain the factorization

x4 + 2x2 − 15 = (x+√

3)(x−√

3)(x2 + 5).

Homework 12.5: Factor these ones:

(a) x3 − 5x2 − 4x+ 20(b) 5x3 + x6 − 14(c) 81x4 − 3x(d) x6y6 − x3y3(e) (x+ y)4 − 16

For more practice, try problems 1-68 of Chapter 5.6.

13. Rational Expressions

In this part of the notes we cover sections 6.1 and 6.2 of the book.

13.1. Why we factor: to simplify rational expressions. A rational expression is

anything of the form p(x)q(x)

, where p(x) and q(x) are polynomials. You already know the

rule for multiplying fractions, it’s as easy as it gets:

p(x)

q(x)· f(x)

g(x)=p(x)f(x)

q(x)g(x).

Also, the rule for canceling fractions is easy to understand

p(x)q(x)

p(x)r(x)=q(x)

r(x).

What’s harder is to know when not to cancel. Beginners in algebra are often tempted tocancel any pair of identical expressions that occur in the numerator and denominator ofa fraction, even though this is almost always the wrong thing to do.


Example 13.1. This is bad:

x2 + 2x+ 1

x2= 2x+ 1.

So is this:x+ 2

x− 1=

2

−1.

And so is this:x2 + x− 1

x− 1= x2.

No, you can only cancel common FACTORS out from the top and bottom of a rationalexpression!

BIG FACT TO REMEMBER FOREVER: You can only cancel common FACTORSfrom the top and bottom of a fraction.

This, then, is one of the biggest reason we had to learn how to factor polynomials inthe previous section. We need to do it in order to perform cancellations that simplifyrational expressions.

Example 13.2. We can simplify the rational expression x2+2x+1x+1

, but we have to factor

the top first to get (x+1)(x+1)x+1

= x+ 1.

It is also important to know when a rational expression cannot be simplified further.This happens when you fully factor the bottom and the top but don’t see any commonfactors.

Example 13.3. The top and bottom of the expression x2−4x2+2x+1

can be factored as (x+2)(x−2)(x+1)2

,

but since there are no common factors, this doesn’t lead to any simplifications. Thereforeit is already fully simplified as it is.

Homework 13.1 Fully simplify these expressions, or say that they are already fullysimplified:

(a) x+2x2−x−6

(b) x+32x+5

(c) x3−1x2−1

(d) x2−9x2−x−6 ·

x2+5x+6x2+x−6

Exercises 27-72 from 6.1 are all good practice for this skill. I must stress here especiallythat, even though I have only given you four homework problems on it here to turn in,you should do a lot more on your own!

Another thing I want to point out is that the book has a lot of exercises where it asksyou to divide polynomials such as this:

x+ 5

7÷ 4x+ 20

9.

This is a stupid kind of problem because nobody who does real math in their job everuses the ÷ symbol. To turn this stupid problem into a worthwhile problem, just flip the


top and bottom of the rational expression on the right and multiply instead to get

x+ 5

7· 9

4x+ 20.

This expression means the same thing as the one with the ÷ symbol above. Now you’vegot a whole new list of good problems (73-90 of Section 6.1) to practice with.

13.2. Adding and Subtracting Fractions. Recall that we discussed fraction additionon the very first day of class, back when we were all young and didn’t know much of thehard things in life, like factoring. Here’s the move you can do with fraction addition:

p(x)

q(x)+r(x)

q(x)=p(x) + r(x)

q(x).

The following move is bad:

p(x)

q(x)+r(x)

s(x)=p(x) + r(x)

q(x) + s(x).

And so is this:p(x)

q(x) + r(x)=p(x)

q(x)+p(x)

r(x).

Again, the key is to show restraint. But you might be wondering:

HOW DO I ADD TWO TERMS WHEN THE DENOMINATORS AREN’T THE SAME?

Answer:

MAKE THEM THE SAME!

More specifically, you need to multiply the top and bottom of each term to get acommon denominator, just like we went over on the first day. The difference this timearound, however, is that we can factor the denominators first, and this will shorten ourwork a great deal when there are common factors.

Example 13.4. Suppose you have to add these: 2x2−1 + x−1

x2−2x−3 . We need to make acommon denominator before we add, and we’d like to make it as small as possible. Thebest way to do this is to start by factoring both denominators before making the commondenominator

2

x2 − 1+

x− 1

x2 − 2x− 3=

2

(x+ 1)(x− 1)+

x− 1

(x+ 1)(x− 3).

In factored form, we can now see that the smallest common denominator is (x + 1)(x −1)(x− 3). So we continue thus:

=2

(x+ 1)(x− 1)

(x− 3)

(x− 3)+

(x− 1)

(x+ 1)(x− 3)

(x− 1)

(x− 1)=

2(x− 3) + (x− 1)(x− 1)

(x+ 1)(x− 1)(x− 3).

From here on out it is just a bunch of polynomial multiplication and addition to get thefinal answer

x2 − 5

x3 − 3x2 − x+ 3.


Homework 13.2: Add the following fractions, make sure your result is in fully simplifiedform.

(1) y2

y2−25 + 25−10yy2−25

(2) 72x2

+ 4x

(3) 3xx2−25 −

4x+5

(4) 3y+7y2−5y+6

− 3y−3

(5) 2x+1x2−yx+6

− x+3x2−5x−6


SPECIAL ANNOUNCEMENT: The first gateway exam will consist of a randomchoice of five excercise from that list.

14. Polynomial Division

In this part of the notes, we cover Sections 6.4 and 6.5 of the book. Unfortunately thewritten part of these notes will be very short because it is extremely time consuming todo a good job of formatting polynomial long division. So I highly recommend reading therelevant sections 6.4 and 6.5 of the book which explain how it works. Here I will simplyexplain the concept of what we are doing and list your homework exercises. All workedexamples will be done in class, on the board.

14.1. Facts to remember about the division of whole numbers. If I asked you todivide 15 by 3, you would say the answer is 5. This is because

15 = 3 · 5.

If I asked you to divide 16 by 3, and give me the answer in “remainder” form, you wouldsay that the answer is 5 with a remainder of 1, and this is because

16 = 3 · 5 + 1.

More generally, if I asked you to divide a by b, and the answer you gave was “c, with aremainder of r,” then this would mean that

a = b · c+ r,

where r < b.

Homework 14.1: Perform the following whole number divisions, giving whole numberanswers with remainders. Then write the corresponding a = b · c + r equation out toremind yourself of what your answer means.

(a) 16/2(b) 16/3(c) 16/4(d) 16/5(e) 16/6

Now remember how whole number long division works. I will write some examples onthe board. Now you do these.


Homework 14.2: Perform these long divisions. Write your answer in “whole number +remainder” form.

(1) 7)712

(2) 14)5, 821

(3) 213)8, 523, 755

14.2. Dividing polynomials. Dividing polynomials (with remainder) is exactly thesame as dividing numbers. If I say, “dividing p(x) by q(x) gives an answer of d(x) withremainder r(x)” then this means that p(x) = q(x)d(x) + r(x), where the highest term ofr(x) is lower than the highest term of q(x). For example, if you divide x3 + 2x2 + x + 1by x2 + 1, the answer you will get is x+ 2 with a remainder of −1. This means that

x3 + 2x2 + x+ 1 = (x2 + 1)(x+ 2)− 1.

Or, if we want to rewrite things in fraction form, we could say that

x3 + 2x2 + x+ 1

x2 + 1= x+ 1− 1

x2 + 1.

The key to understanding how polynomial long division works is to remember thatit is exactly the same as long division of whole numbers, if you remember how decimalnotation works. Remember, writing

57, 428

is really just short hand for

50, 000 + 7, 000 + 400 + 20 + 8 = 5 · 104 + 7 · 103 + 4 · 102 + 2 · 101 + 8 · 100.

To understand how polynomial long division works, it is helpful to do whole number longdivision with all the numbers written in the form above. Polynomial long division is verysimilar, but with x replacing 10. I’ll do a couple of examples, here are some polynomiallong division problems of your own to do:

Homework 14.3: Perform these long divisions. First write them in “answer with re-mainder” form, then write the answer out in “fraction” form. A problem like this is quizeligible, so for more practice try problems 1-50 of Section 6.4 in the book.

(a) (28x3 − 7x2 − 16x)÷ (4x2)(b) (5x3 + 2x+ 1)÷ (2x+ 4)(c) (x2 + 3x− 10)÷ (x− 2)(d) (9x3 − 3x2 − 3x+ 4)÷ (3x+ 2)(e) (3x3 − x3 + 4x2 − 12x− 8)÷ (x2 + 2)(f) (x4 − x3 − 7x2 − 7x− 2)÷ (x2 − 3x− 2)(g) (x4 − x2 + 1)÷ (x2 + x+ 1)

14.3. Synthetic division. If you divide any polynomial by a polynomial of the formx − a, it always goes in the same, nice simple way. Synthetic division is a method ofdividing polynomials in this form that “cuts the fat.” I’ll do an example. You nowpractice on these problems:Homework 14.4: Use synthetic division to perform these divisions. Then write yourfinal answer out in simplified “fraction” form. Try problems 1-18 of Section 6.5–these arequiz eligible.


(a) (x2 + x− 2)÷ (x− 1)(b) (5x3 − 6x2 + 3x+ 11)÷ (x− 2)(c) (5x2 − 12x− 8)÷ (x+ 3)(d) (x4 − 245)÷ (x− 4)

14.4. A trick for quickly computing the value of a polynomial at a given num-ber. Suppose p(x) = q(x)(x − a) + r. Then p(a) = q(a) · 0 + r. Therefore the value ofp(x) at a is just the remainder you get when you divide p(x) by x − a. Using syntheticdivision, this is often a lot faster to find then plugging in for a. Use it one these problems.

Homework 14.5: Use synthetic division to compute the following things. For morepractice, try 19-32 of section 6.5. These are quiz-eligible.

(a) f(4), where f(x) = 2x3 − 11x2 + 7x− 5(b) f(3), where f(x) = x4 + 5x3 + 5x2 − 5x+ 6(c) f(−4), where f(x) = 3x3 + 7x2 − 22x− 8

15. Solving Polynomial and Rational Equations

In this part of the notes we cover a little of Section 5.7 of the book, then move ontoSection 6.6.

15.1. Solving equations involving polynomials. Up to now, we’ve been focused pri-marily on solving equations which can be reduced to something of the form x = k,eventually. However, if higher powers of x occur in an equation, you might not be able tosimplify it so much. For example, consider the equation

2x2 − 3x+ 4 = x2 + 2.

Unlike most of the equations we’ve dealt with so far, you can’t reduce it to the simpleform x = k at the end, and the first step to solving an “polynomial looking” equation likethis is to stop trying to make it look like x = k. Instead, you have a new goal.

New Goal: When faced with a polynomial-looking equation, first try to get it in the formp(x) = 0.

So, for example, the equation above can be changed into the following form

x2 − 3x+ 2 = 0.

Why do this? Well notice that, if we factor the left side this equation it becomes

(x− 2)(x− 1) = 0.

It is now easy to see that this equation will only come out true if x = 2 or x = 1(notice also, that we could have just applied the quadratic formula right away to findthese solutions, without having to factor). Similarly, solving any polynomial equationwill involve these same steps:

(1) Rewrite the original equation in the form p(x) = 0(2) Factor p(x) to get something of the form k(x− a)(x− b)(x− c) · · · (x− d) = 0(3) Read off the answers from this factored form.

Homework 15.1: Find all solutions (if any) to the following equations.


(a) x2 − 4x = 45(b) 3x2 = 5x(c) x(x− 3) = 18(d) (x+ 1)2 = x− 2

For more practice, try problems 1-46 of Section 5.7 of the book.

Before moving on, I want to end with a warning example.

Example 15.1. Let’s find all solutions to x3 − 2x2 + x − 5 = x2 − 5 − x3. If we firstcancel the 5’s, this turns into

x3 − 2x2 + x = x2 − x3.

Since we see x’s on all remaining terms, it’s tempting to just divide both sides by x to get

x2 − 2x+ 1 = x− x2,

which reduces to

2x2 − 3x+ 1 = 0.

Factoring this gives

(2x− 1)(x− 1) = 0

so it seems we have two solutions, x = 12

and x = 1. But wait, there is another solutionto the original equation which we have missed: x = 0! We don’t discuss this often, butwhenever you divide by the variable x, you are assuming that x is not 0, otherwise it’sillegal to divide. You also run the risk of eliminating the solution x = 0 as we did here.So remember: ALWAYS CHECK THAT x = 0 IS NOT A SOLUTION BEFORE YOUDIVIDE BY x!

15.2. Solving equations involving rational functions. There is one extra step tosolving an equation involving rational functions: clearing denominators. This is just likehow it is when you have fractions involving constant numbers, except now there may bepolynomials in the denominator.

Example 15.2. Let’s solve 5 + 23x

= 2x−1x

+ 1x2

. The first thing we need to do is multiplyboth sides of this equation by the least common multiple of all the denominators appearingin this equation. In this case, the least common multiple is 3x2, so we get

15x2 + 2x = 6x2 − 3x+ 3.

Having reduced this equation to a polynomial equation, we know what to do. Puttingeverything on one side turns it into

9x2 + 5x− 3 = 0.

Now, instead of factoring using one of the primitive methods, I’m just going to plug thisbaby into the quadratic formula to get the solutions

−5±√

133

18.

Good thing I did that, because in this case only the quadratic formula was going to dothe trick.


But this example was easy. What happens if we have real polynomials occurring in ourdenominators? The trick, as always is: SAVE TIME BY FACTORING DENOMINA-TORS FIRST!

Example 15.3. Let’s solve the equation

9

x2 − 9+

4

x+ 3=

2

x− 3.

If we didn’t factor, we might think that the least common multiple is (x2−9)(x+3)(x−3),which is a real nightmare to multiply out. However, if we factor first, we get

9

(x+ 3)(x− 3)+

4

x+ 3=

2

x− 3.

So now it is obvious that the LCM is actually just (x+ 3)(x− 3), which is much shorter.Multiplying both sides of the equation by this LCM gives

9 + 4(x− 3) = 2(x+ 3).

We now do the ancient dance to get x = 5.

Homework 15.2 Solve for x.

(a) x+ 7x

= −8

(b) 5x+4

+ 3x+3

= 12x+19x2+7x+12

(c) 2x+3− 5

x+1= 3x+5

x2+4x+3

(d) 2x3−1 + 4

(x−1)(x2+x+1)= − 1

x2+x+1


I’ll finish this section with one final remark. Remember that a rational function p(x)q(x)

is always undefined when q(x) = 0. Therefore, any equation involving p(x)q(x)

will not make

sense when q(x) = 0. However, sometimes when you go through the process above, youend up with answers which will make some q(x) = 0, and such answers will NOT countas solutions to the original equation. So: ALWAYS CHECK YOUR ANSWERS AT THEEND TO MAKE SURE THEY ARE DEFINED IN THE ORIGINAL EQUATION!

16. Radical expressions, Part I

In this section of the notes we cover sections 7.1–7.3 of the book.

16.1. Definition and elementary properties. If n is an even number, and x is apositive number, then there are always two numbers y such that yn = x, one negative,and one positive. By definition, n

√x is the positive number y such that yn = x. Also n

√x

is undefined when x < 0.

Example 16.1. Both 24 = 16 and (−2)4 = 16. However 4√

16 is, by definition, alwayspositive, so 4

√16 = 2. On the other hand, 4

√−1 is undefined, because no matter what

number y that you choose, y4 will come out positive.

This means that nth roots have the following interesting property when n is even:

Proposition 16.2. If n is an even number, then n√xn = |x|.


Example 16.3. If you substitute (−2) for x and n = 4 in the equation above, you get4√

(−2)4 = 4√

16 = 2, which is | − 2|.

One the other hand, if n is an odd number, then every number x has an nth rooty = n

√x such that yn = x. Moreover, y will be unique, and it may be negative. This

makes them a little bit more simple to work with, algebraically, because if n is odd, thenn√xn = x for all x.

Example 16.4. 33 = 27, but (−3)3 = −27. Therefore, unlike in the case with even roots,there is no choice to be made in the definition of 3

√27. Also, notice that 3

√−27 = −3, so

negative numbers do have cube roots, and these will be negative.

Homework 16.1 Simplify (or compute) each of the following.

(a)√

(−4)2

(b)√x2 − 2x+ 1

(c) 3

√−8125

(d) − 7√−1

(e) 5√−32(x− 1)5

(f)3√

4√

16 +√

625

For more practice, try Exercises 33-88 of Section 7.1.

16.2. Exponential notation. Using radicals, it is now possible to define rational expo-nents.

Definition 16.5. Suppose p and q are whole numbers. Then xp/q = q√xp. In particular,

x1/q = q√x.

Just like we defined negative exponents the way we did in order to make the lawxpxq = xp+q work, this definition makes sure that the law (xp)q = xpq works (at leastassuming x ≥ 0). All the other laws for exponents work with this definition as well.Therefore, doing algebra with radicals can always be reduced to laws of exponents. Soif you’re comfortable with exponents (as you should be) it’s a good idea to start thesimplification of a complicated radical expression by converting the radicals into exponentsfirst. And in any event, you need to be able to convert from rational expressions to radicals,in order to actually compute values.

Example 16.6. Let’s simplify ( 3√xy)18. First, we convert to exponent notation to get

((xy)1/3)18.

Applying the exponent rule (xp)q = xpq, we get

(xy)6.

Now we distribute the 6 back in to get our final answer

( 3√xy)18 = x6y6.

Example 16.7. Let’s simplify7√x2 ·√x. We begin by putting the first expression into

exponent notation to get

(x2)1/7x1/2.


Applying the rule (xp)q turns it into

x2/7x1/2,

and using the rule xnxm = xn+m turns it into

x2/7+1/2.

Since 2/7 + 1/2 = 4/14 + 7/14 = 11/14, this becomes

x11/14,

turning this expression back into radical notation gives us our final answer:

7√x2 ·√x =

14√x11.

Homework 16.2 Compute the following expressions, or reduce them to radical notationso that no fractional exponents appear.

(a) −161/4

(b) (3xy3)1/5

(c) 10002/3

Now go the other way: rewrite each of these in exponent notation.(d) 5√

13x

(e) 2x 3√y2

Now simplify the following expressions as much as possible, so that no negativeexponents appear, at most one constant appears (possibly under a radical sign)and each variable occurs at most once.

(f) 334 3

14

(g) (125x9y6)1/3

(h) (2y1/5)4

y3/10

(i) 5√x15y20

(j) 4√x2y6

(k)√

3 · 3√

3

(l)3√y2

6√y

(m)√√

x2y

For more practice, try problems 1-112 of Section 7.2 the textbook.

17. Radical Expressions, Part II

This part of the notes covers 7.3, 7.4, and little bit of 7.5 from the textbook.

17.1. Multiplying and Dividing radicals. Radicals actually behave quite nicely whenit comes to multiplication and division, in particular we have the following two rules:

Mult: n√f(x) · n

√g(x) = n

√f(x) · g(x)

Div:n√f(x)

n√g(x)

= n

√f(x)g(x)


The exponent notation makes these rules easy to remember. In exponential notation,the multiplication rule translates out to

f(x)1/ng(x)1/n = (f(x)g(x))1/n,

and the division rule translates to

f(x)1/n

g(x)1/n=

(f(x)

g(x)

)1/n

.

Example 17.1. Let’s simplify4√

6x2 · 4√

8x2. First we use the multiplication rule to puteverything under a single radical, to get

4√

6x2 · 8x2.

Now we simplify the radicand to get

4√

48x4.

We now apply the radical to each part of the newly simplified radicand to get

4√

48 · 4√x4 =

4√

48 · |x|.

Finally, notice that 48 = 3 · 16 = 3 · 24 (one way to see this is to keep on dividing 48 by

2), so 4√

48 =4√

24 · 3 =4√

24 · 4√

3 = 2 4√

3. This leaves us with our final answer

4√

6x2 · 4√

8x2 = 24√

3|x|.

Homework 17.1: Simplify the following radicals as much as possible.

(a)√

2x3·√

32

(b) 3√

81x8y6

(c)√

8x4

9z2y6

(d) 3

√x4

8y3

(e)√200x3√10x−1

(f)√72xy

2√2

Try the exercises of Section 7.3, as well as 29-62 of Section 7.4 for more practice.

17.2. Polynomials under the radical. When you have polynomials under a radical, itis important to practice restraint. Just as it is hard to deal with addition and subtractionunder a fraction bar, so you have to be very careful about addition and subtraction underthe radical. For example, the following rule is NOT TRUE :

n√f(x)± g(x) = n

√f(x)± n

√g(x).

In fact, the only way to deal with addition and subtraction under the polynomial is...you guessed it, FACTORING! Although usually there really is nothing at all that willwork, sometimes when you factor you might turn something like this

√f(x) + g(x) into

something that looks like this√

[h(x)]2p(x), which simplifies down to |h(x)|√p(x).


Example 17.2. Let’s simplify√

3x− 6 ·√x2 + x− 6. We cannot just apply the radical

to each term. However, we can multiply everything under a common radical to get√(3x− 6)(x2 + x− 6).

Remember those parentheses! Now at this point you might be tempted to multiply (3x−6) ·(x2+x−6) out, but that will only take us farther from our one true shot at simplifyingthis. Instead we factor (3x− 6) = 3(x− 2), and (x2 + x− 6) = (x− 2)(x+ 3), to get√

3(x− 2)2(x+ 1).

Applying the radical to each factor now gives our final answer of

|x− 2|√

3x+ 3.

Homework 17.2: Simplify the following expressions.

(a) 3√

2 · 3√

4x+ 4 · 3√−x4 − x3

(b)√

50x− 25 ·√

6x2 + 5x− 5(c) 3√x3 − 1 · 3

√x2 − 2x+ 1

(d)3√a3+b33√a+b

(e)√x3−1√

x3+x2+x

Similar problems to these occur also in Sections 7.3 and 7.4 of the book’s homework.

17.3. Addition and Subtraction outside of the radical. Because taking radicals issecretly just a form of exponentiation, it works out well when you’re just multiplying anddividing a bunch of things together. When you try to add fractions together, however,you have to be very careful! The only general rule of radical addition is just a specialcase of the distributive law:

f(x) n√g(x) + h(x) n

√g(x) = (f(x) + h(x)) n

√g(x).

Example 17.3. You can simplify 2x√

3y3 + 3x√

3y3 down to (2x+ 3x)√

3y3 = 5x√

3y3.

However, you can’t do diddly to 2x√

3y+3x√

3y3, nor can you do anything with 2x 4√

3y3+

3x 5√

3y3, because the radical parts of each term need to be identical before you can addthem together.

Homework 17.3: Simplify as much as possible.

(a) 8√

5 + 3√

5(b) 8

√17− 5

√19− 6

√17 + 4

√19

(c) 7√

7− 3√x+ 6 3

√7 + 5 3

√x

Sometimes, however, by simplifying terms you can make them have a common radicalpart.

Example 17.4. At first glance, it looks like we cannot combine 8√

45x3 +√

5x down to asingle term, because they have different radical parts. However notice

√45x3 =

√9x25x =

3|x|√

5x. So we get

8√

45x3 +√

5x = 24|x|√

5x+√

5x = (24|x|+ 1)√

5x.

So now we can some problems with a little more meat on the bones.

Homework 17.4: Simplify as much as possible.


(a) 3 3√

24 + 3√

81

(b) 3√

54xy3 + y 3√

128x(c)√

4x− 12 +√x− 3

And finally, for a little bit of extra juice, have a glass of this:

(d) 7√

2x3 + 40x3√150x2

5x2√3x

Section 7.4 of the book has a bunch more problems like this.

18. Radical Expressions, Part III

Today we cover Sections 7.5 and 7.6 from the book.

18.1. Rationalizing denominators: the conjugate trick. Multiplying radical expres-sions together works just like polynomial multiplication at first. Then you simplify.

Example 18.1. To multiply (3√

7− 2√

3)(√

5 + 5√

17) out, we start by foiling just as wewould with two polynomials:

3√

7 ·√

5 + 15√

7 ·√

17− 2√

3 ·√

5− 10√

3 ·√

17

Now we just simplify each term using the rules for radicals. Here there isn’t much to do,and we get:

3√

35 + 15√

119− 2√

15− 10√

51.

Without further ado, let’s practice.

Homework 18.1: Simplify the following expressions.

(a) (5 +√

2)(6 +√

2)(b) (7− 2

√7)(5− 3

√7)

(c) (4−√x)(3−

√x)

(d) (x+ 3√y2)(2x− 3

√y2).

For more practice, try problems 1-38 of Chapter 7.5 in the book.

Rationalizing a denominator just means getting rid of a radical in the denominator likeso:

5√

7√3

=5√

7√3·√

3√3

=5√

21

3.

This example is easy, but sometimes it is harder to rationalize a denominator.

Example 18.2. Let’s rationalize 6√5+√3. The trick here is to multiply the top and the

bottom by something a bit more complicated:

6√5 +√

3·√

5−√

3√5−√

3=

6(√

5−√

3)√5 ·√

5−√

5 ·√

3 +√

3 ·√

5−√

3 ·√

3=

6(√

5−√

3)

5− 3= 3(√

5−√

3).

Here is the trick that got Example 18.2 going: we multiplied the top and bottom of thefraction by the conjugate of its denominator. Given any expression of the form A + B,its conjugate is the expression A−B. And on the other hand, the conjugate of A−B isA+B. The key point about conjugates is that

(A−B)(A+B) = A2 −B2


which is nice if A and B are both square roots.

Homework 18.2: Rationalize the denominators of the following expressions, then sim-plify as much as possible.

(a) 9√3y

(b) 103√16x2

(c) 54√x2y7

(d) 13√11−3

(e)√b√

a−√b

(f)√x−2√x−5

(g)2√x+√y

√y−2√x

For more practice, try problems 39-92 of Chapter 7.5

18.2. Solving equations involving radicals. Now we are ready to solve equationsinvolving radicals. The basic idea is to try to reduce it to the form

n√f(x) = g(x).

Taking the nth root of both sides will then turn it into

f(x) = [g(x)]n

, which will be a polynomial equation you know how to solve. But beware: if n is an evennumber, the second equation might have more solutions than the first, which means thatyou should always check your answers at the end to make sure they actually work.

Example 18.3. Let’s solve x − 2√x− 3 = 3. The first step is to isolate the radical on

one side, which gives usx

2− 3

2=√x− 3.

Now we square both sides to get (x

2− 3

2

)2

= x− 3.

Multiplying the left side out then gives us

x2

4− 3

2x+

9

4= x− 3.

We can now multiply both sides by 4 to get

x2 − 6x+ 9 = 4x− 12.

Using ordinary polynomial methods, we then get

x2 − 10x+ 21 = 0.

Factoring the left side gives (x− 7)(x− 3) = 0. So we get two solutions: x = 3 and x = 7.When we plug them in, they both work.

Homework 18.3: Solve for x.

(a) x =√

7x+ 8


(b) 3√

2x− 6− 4 = 0(c) 2

√4x+ 1− 9 = x− 5

Sometimes it takes a bit more work to simplify a radical equation, and the basic processhas to be repeated twice.

Example 18.4. Let’s solve√x− 7 = 7 −

√x. To begin with, we move

√x to the left

side to get √x− 7 +

√x = 7.

Now we have the radicals on one side, and everything else on the other, but unfortunatelythere is no nice way to combine these two terms into a single radical. When we squareboth sides, we get

(√x− 7)2 + 2

√x ·√x− 7 + (

√x)2 = 49

This doesn’t look like much progress, but after simplifying there is only a single radicalterm left over

x− 7 + 2√x(x− 7) + x = 49.

Now we use the usual process, rewriting the equation in the form√x(x− 7) = −x+ 28

. Squaring both sides again turns it into x(x−7) = x2−56x+ 784, which ultimately onlyhas the solution x = 16.

Homework 18.4: Solve for x:

(a)√

2x− 5 =√x+ 4

(b)√x− 4 +

√x+ 4 = 4

(c)√x+ 2 +

√3x+ 7 = 1

For more practice on this and Homework 18.3, try probems 1-38 of Section 7.6.

19. Quadratic Functions

Today we cover sections 8.1-8.3 of the book.

19.1. Completing the square. If you add the number(b2

)2to an expression of the form

x2 + bx, you get

x2 + bx+

(b

2

)2

=

(x+

b

2

)2

.

This process is called completing the square.

Example 19.1. Let’s complete the square for the expression x2− 10x. Here b = −10, so(b2

)2= (−5)2 = 25. So, completing the square, we get x2 − 10x+ 25 = (x− 5)2.

Homework 19.1: Complete the square.

(a) x2 + 2x(b) x2 − 6x(c) x2 − 1

3x

(d) x2 + 95x



WARNING: Completing the square changes the algebraic expression you are workingwith, so you can’t just do it will nilly!

19.2. Completing the square in order to solve quadratic equations. One use ofcompleting the square is to solve equations of the form ax2 + bx + c = 0 (or equationswhich can be rewritten in this form). The basic idea is to first rewrite the equation inthe form x2 + bx = c, then use completing the square to rewrite the equation in theform (x + C)2 = K, then take square roots to get x + C = ±

√K, which then gives

x = −C ±√K.

Example 19.2. Let’s use completing the square to solve x2 + 3x− 1 = 0. We’ll start byrewriting in the form

x2 + 3x = 1.

We can complete the square on the right side by adding(32

)2= 9

4. However, whatever we

do to one side of an equation, we must do to another. So we add 94

to both sides to get

x2 + 3x+9

4= 1 +

9

4.

Factoring the left side, and simplifying the right side, gives us(x+

3

2

)2

=13

4.

Taking square roots gives

x+3

2= ±

√13

4.

Subtracting 32

from both sides gives us our solutions

x = −3

2±√

13

2.

If you plug it in, you’ll notice that this gives exactly the same answer as the quadraticformula. In fact, if you remember the proof I gave in class, it started by completing thesquare.

Homework 19.2: Solve the following equations by completing the square.

(a) x2 + 4x = 32(b) x2 − 8x+ 1 = 0(c) 3x2 − 6x+ 2 = 0(d) 8x2 − 4x+ 1 = 0


19.3. Using completing the square to graph quadratic equations. Completing thesquare can be used to graph quadratic functions of the form f(x) = ax2 + bx + c. Theyall have the same basic “bowl” shape we are familiar with, although to place it correctlywe must rewrite the function in the form

f(x) = a(x− k)2 + h


where h and k are constants. In this form, we can read off the following facts about thegraph of f(x).

• The vertex (bottom of the bowl) occurs at the point (k, h).• If a is positive, then the bowl faces up, and if a is negative, then the bowl faces

down.

I will explain why this is true in class. Using these two pieces of information, togetherwith our knowledge of the x- and y-intercepts of the function, we can draw a pretty goodlooking graph. I’ll show you some examples in class.

Homework 19.3: Sketch graphs of the following equations.

(a) f(x) = (x+ 1)2 − 1.(b) f(x) = 2(x− 3)2 + 1.

(c) f(x) = 54−(x− 1

2

)2.

Completing the square comes into play when we turn an expression of the form ax2 +bx + c into one of the form a(x − k)2 + h. The way to do this is to complete the squarefor the ax2 + bx part of the expression first, like so:

ax2 + bx+ c = a

(x2 +

b

ax

)+ c = a

(x2 +

b

ax+

(b

2a

)2

−(b

2a

)2)

+ c

= a

((x+

b

2a

)2

− b2

4a2

)+ c = a

(x+

b

2a

)2

+ c− b2

4a

The upshot of this little derivation is that, if you start with ax2 + bx+ c and rewrite itin the form a(x− k)2 + h, then

• k = − b2a

• h = c− b2

4a

If you want, you can memorize these formulas for k and h or you can use completingthe square to find them as you need.

Example 19.3. Let’s rewrite f(x) = −2x2 + 8x− 1 in the form a(x− k)2 + h. The firststep is to factor out the −2 from −2x2 + 8x to get

−2(x2 − 4x)− 1.

The next step is complete the square for x2 − 4x by adding(−4

2

)2= 4 to it. However,

remember we cannot change the value of the expression, so we must immediately subtract4 as well, like so:

−2(x2 − 4x+ 4− 4)− 1.

The advantage of this seemingly pointless move is that we can rewrite x2 − 4x + 4 as(x− 2)2, and get

−2((x− 2)2 − 4)− 1.

We now simplify to getf(x) = −2(x− 2)2 + 7.

We are now ready to graph this function.

Homework 19.4: Graph these.


(a) x2 − 2x− 3(b) 2x2 + 4x− 3(c) 2x− x2 − 2

20. Quadratic functions and word problems.

This section discusses word problems from Section 8.3 in the book, which are associatedwith quadratic functions. Like systems of linear equations, our knowledge of quadraticequations allows us to draw surprisingly powerful conclusions about real world problems.

20.1. Motion under gravity. If, at time t = 0, the object is fired at a rate of V metersper second from an initial height H, then its height h(t) at all future times t will be givenby the equation

(20.1) h(t) = −g2t2 + V t+H

Here, t is time in seconds, h is height in meters, and g ≈ 9.8m/s2 is the gravitationalacceleration constant for objects close to the earth (on other planets it is different). Ofcourse, the equation becomes invalid as soon as some other force acts on the object, orit hits the ground (and even if none of these things happen, we are ignoring the effect ofwind resistance, which in real life can make quite a difference).

It is an amazing fact that, using this equation and our knowledge of the object’s state attime t = 0, we can predict its future course with such precision. Not only that, but givenour geometric knowledge of what the graph of a quadratic function looks like, we canpredict important events in the object’s motion. In particular, using the vertex formulawe learned last time, we can discover that the vertex of the graph of h(t) will always occurat the point

(k, h) =

(V

g,H +

V 2

2g

).

Here, our horizontal axis is the t-axis and it represents time, and our vertical axis is theh axis and represents the height of the object. Since the coefficient −g

2of t2 is negative,

the graph of h(t) will face down, and the vertex will represent the highest height that itreaches. Thus, this statement:

The vertex of the graph of h(t) occurs at the point(Vg, H + V 2

2g

).

translates to the following physical fact:

In the absence of other forces, an object will attain a maximum height of H + V 2

2gmeters

Vg

seconds after it is launched from the height H meters with velocity V meters/second.

Example 20.1. An object is fired from the ground straight up into the air at a rate of20m/s. How high will it reach?

Solution: We can set ground level = 0, so that the initial height is H = 0. MeanwhileV = 20. Plugging this into the formula above, we deduce that the object will reach amaximum height of 202

2g= 200

g≈ 20.4 meters. Moreover, even though the problem doesn’t

ask for it, we can easily also calculate that it will take the object a little over 2 secondsto reach that height.


Incidentally, for those of you who prefer miles per hour, that initial speed is about 44miles per hour, and the height reached will be a little over 60 feet. This means that if youcan get your car traveling 44 mph straight up a wall somehow (like batman), it wouldkeep flying another 60 feet up in the air on its own after you left the wall.

Not only can we predict how long it will take for the object to reach its maximumheight, we can predict how long it will take the object to reach any height between itsstarting height and maximum height using the quadratic formula! Namely, the followingmathematical fact follows from the quadratic formula:

The function −g2t2 + V t+H = N has solutions t =

−V±√V 2+2g(H−N)

−g .

And this translates into the following physical fact:

In the absence of other forces, an object will attain a height of N metersV±√V 2+2g(H−N)

g

seconds after it is launched from an initial height of H meters with velocity V m/sec.

Example 20.2. You drop a marble off the top of a building that’s 52 meters tall. Howlong will it take to hit the ground below?

Solution: Here our initial height is H = 52, however since we are simply dropping themarble, the initial velocity is 0. When the object hits the ground, it has height N = 0.Plugging all of this in the formula above, we see that the marble will hit the ground

0±√

02 + 2g(52− 0)

g= ±2

√26g

g

seconds later. Throwing out the negative number (which makes no physical sense), andplugging this into the calculator with g = 9.8, we see that the marble hits the groundabout 3.13 seconds later.

Homework 20.1 A person stands at the edge of a 120 meter tall cliff and throws abaseball down it at a speed of 40 meters/second.

(a) Find the equation that defines the height of the ball h(t), t seconds after it wasthrown down the hill.

(b) How long will it take for the ball to make it halfway down the cliff?(c) How long will it take for the ball to reach the bottom of the cliff?

For more practice, try problems 57 and 58 of section 8.3. The following problem is abit more tricky but you need to understand it for tests and quizzes.

Homework 20.2 From the ground, you throw a ball straight up in the air, and exactlyone second later it hits a flag 10 meters in the air. The flag stopped the ball’s motionimmediately, and it fell back to the ground.

(a) How fast did you throw that ball?(b) How long did it take the ball to get halfway to the flag?(c) How high in the air would the ball have gone if the flag was not there?(d) How long was the ball in the air before hitting the ground again?(e) How long would the ball have been in the air if it had not hit the flag?


20.2. Optimization. The basic idea of optimization is to find the maximum (or mini-mum) value of some function f(x). Using the vertex formula, we can do this for quadraticfunctions. This is what we did in Example 20.1: we found the “optimal” height usingour knowledge of quadratic functions. The following is a more complicated example of anoptimization problem.

Example 20.3. You have 100 feet of fencing, and would like to use it to build a rect-angular enclosure, one side of which lies against a river (and therefore needs no fencing).You wish to enclose as much area as possible, what should your dimensions be?

Solution: Here the quantity you are trying to optimize is enclosed area: you are trying tomaximize it. Let A denote the area. You have control over the dimensions of the fence,x and y, although this is subject to the restraint that x + 2y = 100, since you only have100 feet of fencing to work with (and we should use it all if we want to make the area asbig as possible).

In terms of the variables x and y, we know that the area of the fence is given by

A = xy.

Our strategy is to rewrite this equation so that A defines a quadratic funtion of x. Oncewe’ve done this, we can find the maximum area (together with the dimensions that achieveit) using the vertex formula. The restraint equation x+ 2y ≤ 100 can be rewritten as

y = −x2

+ 50.

Substituting this expression for y in our area formula, we obtain A = −12x2 + 50x. Using

the vertex formula on this expression, we see that the maximum area we can enclose is1250 square meters, which occurs when the width x = 50. This then forces y = 25, whichgives us the full dimensions we should use.

When doing an optimization problem, you want to identify the following things:

• The quantity you are trying to optimize.• The underlying variables which affect the value of the quantity you are trying to

optimize.• Some kind of a restraint on the underlying variables.

In the problem above, the quantity we were trying to optimize was the area enclosedby the fence. The variables which affect the area enclosed by the fence are its length andwidth. The restraint was the condition that the fencing have a total length of 100 feet.

Identifying the quantity you are trying to optimize, and the variables on which itdepends, will allow to you write it as a function. The restraint on the underlying variablescorresponds to an equation which will allow you to write one variable in terms of another.This allows you to write the quantity you are trying to optimize as a function of a singlevariable. In this section, this function will be quadratic, which allows us to find its optimalvalue.Homework 20.3: Again, you are trying to make a rectangular enclosure. This time, oneof the walls must be a brick wall, the other two will be made of wood. The wooden wallscost $2 per foot of length, and the brick wall costs $6 per foot of length. You have $1000to spend on the walls–what’s the biggest area you can enclose?


To get started on this, start by observing that the quantity you are trying to optimizeis again the area to be enclosed, and the variables it depends on are again the length andwidth of the fence. However, this time the restraint is given by the cost of the fence andthe amount of money you have.

Homework 20.4: Problems 67 and 69 of Section 8.3 in the book.

For more practice on the problems discussed in this section, try problems 57–70 inSection 8.3 of the book.

21. Exponential Functions and Logarithmic Functions

In today’s notes we cover Sections 9.1 of the book, about exponential functions andlogarithms.

21.1. Definition of Exponential Functions. An exponential function is a function ofthe form ax, where a is a fixed constant greater than 0. For the sake of concreteness, let’sthink about the function f(x) = 2x. It is easy to figure out the value of f(x) = 2x whenx is a whole number. For example:

• f(0) = 20 = 1• f(1) = 21 = 2• f(2) = 22 = 4• f(3) = 23 = 8• f(−1) = 2−1 = 1

2

• f(−2) = 2−2 = 122

= 14

• f(−3) = 2−3 = 18

And so on. We can also figure out the value of 2x when x is rational number of theform p

qusing the exponent rules. For example:

• f(1/2) = 21/2 =√

2 ≈ 1.41

• f(3/4) = 23/4 =4√

23 = 4√

8 ≈ 1.68

• f(7/8) = 27/8 =8√

27 = 8√

128 ≈ 1.83

• f(15/16) = 215/16 =16√

215 ≈ 1.91

• f(99/100) = 299/100 =100√

299 ≈ 1.99

These, of course, are very hard to actually compute, but they are at least well defined.And notice that they are “well-behaved” in the sense that, as our x values get closer andcloser to 1, f(1) gets closer and closer to f(1) = 21 = 2. So, if we somehow did not knowthe value of 21, we could have guessed that its value would be 2 by looking at the valuescomputed above.

Now consider the case when x is not rational, where rational exponent notation doesnot work. For example, how shall we define 2π? The idea is to pick a sequence of rationalnumbers which get closer and closer to π = 3.1415926 . . ., and define 2π to be the limit ofthe numbers approached in this sequence, like so:

• 23 = 8• 23.1 = 231/10 =

10√

231 ≈ 8.57419• 23.14 = 2314/100 =

100√

2314 ≈ 8.81524• 23.141 = 23141/100 ≈ 8.82135


• 23.1415 = 231415/10000 ≈ 8.82441• 23.14159 = 2314159/100000 ≈ 8.82496

None of these numbers is 2π on the nose, but they are getting closer and closer to theactual value. From these computations, we can comfortably guess that the first threedigits of 2π are 8.82. And in fact, if we plug 2π directly into the calculator we find2π = 8.82497 . . ., so in fact the first five digits we found were already correct.

Homework 21.1: Either calculate the value by hand, or approximate the value like Idid for 2π:

(a) 9−.5

(b) 41.5

(c) 2√2

(d) 3−π

Question: Why isn’t the function (−2)x defined?

Now I’ll graph some of the values of f(x) = 2x on the board. Every exponential func-tion ax always has the same basic shape as 2x. ax is always increasing, with y-interceptalways equal 1, and f(1) = a1 = a.

Homework 21.2: Draw a rough graph of each of the following functions. Mark at least5 points on each explicitly:

(a) 3x

(b)(32

)x(c) 2−x

(d) 5−x − 1(e) −3x + 1

21.2. Some applications of exponential functions. Exponential functions occur nat-urally in nature all the time. In fact, pretty much any physical quantity whose growthrate is a constant multiple of its size is given by an exponential function. For example,let us say that you have a population of bacteria growing in a petri dish, and it doublesevery second. Then if you started with 1000 bacteria at time t = 0 seconds, at time tseconds you will have

P (t) = 1000(2t)

bacteria. So for example, one second later you have P (1) = 1000(21) = 2000 bacteria,which makes sense because it was supposed to double after a second. The 2000 bacteriayou have at t = 1 should double to 4000 one second later, at t = 2, and sure enough weplug in and get

P (2) = 1000(22) = 4000.

Exponential growth is fast. If the bacteria keeps doubling every second, then in 30 secondsyou end up with

P (30) = 1000(230) = 1, 073, 741, 824, 000

bacteria.


Homework 21.3: Every year, the bee population triples in a small town. Assuming thetown started with 2 bees, how many bees will it have in 10 years?

Exponential functions also govern the way a bank account grows. For example, supposea bank promised to pay you 3% interest per month on whatever money you keep in youraccount (this is a fantastic deal by the way). Suppose you put P0 dollars in there at first.Then after the first month, you will have the P0 dollars you started with, plus .03 · P0

dollars in interest, for a grand total of

P0 · 1.03

dollars after month 1. In the second month, you will have the P0 · 1.03 dollars from lastmonth, plus another (P0 · 1.03) · .03 for a grand total of

(P0 · 1.03) + (P0 · 1.03) · .03 = (P0 · 1.03) · 1.03 = P0(1.032)

dollars in the bank. More generally, you will have P (t) = P0 · (1.03t) dollars in the bankt months later.

More generally, when someone says that they are going to charge you an interest rateof r per year, compounded n times per year, they actually mean that they are going topay you an interest rate of r/n every nth of a year. If you put P0 dollars into such anaccount initially, then the amount of money you will have at time t years (assuming youspend nothing), will be given by

P (t) = P0(1 + r/n)tn.

Example 21.1. You invest $50 in a bank account which pays out 6% annually, com-pounded monthly. Assuming you don’t touch it, how much money will you have in 5years?

Solution: The fact that it is compounded monthly means that n = 12 in the formulaabove, and since the interest is 6% that makes r = .06. Therefore the amount of moneyin your account after t years is given by

P (t) = 50(1 + 0.06/12)12t = 50(1.005)12t.

Therefore at t = 5 there will be

P (5) = 50(1.005)12·5 = 67.44

dollars in your account.

Homework 21.4: A bank offers two bank accounts. The first one offers 6.4% interest,compounded monthly, while the second offers 6.5%, compounded semi-annually (twice ayear). For each account, calculate how much money you will have after 10 years if youput $100 in. Which one is the better deal?

22. Logarithms

This part of the notes covers Chapter 9.3 of the book, which is about logarithms. Iknow it’s late in the semester, but we’ve just started rocking.


22.1. Basic definition. The definition of a logarithm is a little bit of a brain twister.

Definition 22.1. Let b and c be positive numbers. Then logb(c) is the number that youneed to raise b to in order to get c out. More briefly, logb(c) is the number m such thatbm = c.

For example, when asked to compute log2(8) what you need to do is figure out thenumber that you have to raise 2 to in order to get out 8. After maybe a couple of tries,you find that if you raise 2 to the 3rd power you get 23 = 8. Therefore log2(8) = 3. Hereare some more examples, together with a verification that the answer I gave is in factright.

• log2(1) = 0 (verification: 20 = 1)• log2(2) = 1 (verification: 21 = 2• log2(16) = 4 (verification: 24 = 16).• log3(1/9) = −2 (verification: 3−2 = 1

32= 1

9).

Homework 22.1: Determine which of the following equations are true and which arefalse:

(1) log3(3) = 1(2) log3(6) = 2(3) log3(9) = 3(4) log3(9) = 2(5) log10(100) = 10(6) log10(100) = 2(7) log10(10000) = 5(8) log2(16) = 4(9) log4(16) = 4

(10) log4(16) = 2(11) log4(1/16) = −2(12) log4(1/16) = 1

2

(13) log4(1/16) = 14

(14) log2(.5) = −1(15) log4(2) = .5(16) log4(.5) = −.5(17) log9(27) = 3

2

(18) log27(9) = 23

So now that you are used to verifying whether or not a logarithm has been computedcorrectly, you should have a good feel for the fact that, when computing logb(c), you arelooking for a number n to raise b to and get out c. You’re starting at the base b, andtrying to get up to c using the right kind of exponent n to launch you there. Notice thatif c is smaller than c, you will be looking for a negative number n to launch b down to c.You should use the following strategy for trying to compute logb(c):

(1) Check for whole number answers by guessing whole numbers m and seeing if one“launches” b onto c, i.e. bm = c. If one of them works, great!

(2) If b has a nice whole number nth root, then see if you can launch n√b to a whole

number power m to get c. If one works, then the answer is mn

.


(3) Give up or plug into a calculator.

Example 22.2. Let’s compute log8(512). We start by guessing 2. But then we check andsee 82 = 64, which is not big enough to get us to 512, not even close. So now, thinkingwe’ve got a long way to go, we guess 4 next. But 84 = 4096, way too big. Finally weguess 3, which is our last chance for a whole number answer. We get 83 = 512 on thenose, perfect. So log8(512) = 3.

Now let’s compute log2(1/64). Since 1/64 is smaller than 2, we are going to have touse a negative number in order to launch 2 down to it. We try −5, but this gives us2−5 = 1

25= 1

32, too small. Next we try −6 and get 2−6 = 1

26= 1/64, like we needed.

Therefore log2(1/64) = −6.

Now let’s compute log8(32). We try 2, but get 82 = 64, already too big. On the otherhand, 1 is clearly not enough to get us there. So there is no whole number answer. Butthis does not mean we have to give up yet! Since 8 has a nice cube root, namely 3

√8 = 2,

there is still hope. And in fact, since 25 = 32, we find that ( 3√

8)5 = 85/3 = 32. Thereforelog8(32) = 5/3.

Now let’s try to compute log2(14). We try 23 = 8 and it’s too small, but 24 = 16 is toobig. So there’s no whole number answer to this one. Unfortunately, the base 2 does nothave a nice whole number nth root of any kind either. So we can’t get a fraction answerout. So we just give up and either leave it written as log2(14) or, if we need to a closersense of its value for some other purpose like graphing, we plug it into our calculator toget log2(14) = 3.807 . . ..

Homework 22.2: Compute the following logarithms by hand! Use the process I outlinedabove. Only give up if you can’t get a fractional answer out.

(a) log4(64)(b) log3(1/27)(c) log√2(8)(d) log25(1/5)(e) log4(32)(f) log8(1/32)(g) log10(100, 000)(h) log10(.0000000001)(i) log100(.1)(j) log9(1)(k) log99999999999(1)(l) log27(1/9)

For more practice, try problems 21 through 42 in Section 8.3 of the book.

22.2. The logarithm as the inverse of the exponential function. After workingwith logarithms a bit, you should be able to reason for yourself that the following state-ment is true.

Proposition 22.3. For any number b > 0 and x > 0, the following statements are true:


(1) blogb(x) = x(2) logb(b

x) = x

Proof. Both statements really do just amount to restatements of the definition of thelogarithm. Since logb(x) is the number n such that bn = x, statement (1) is just what youget when you actually plug in logb(x) for n. It is similarly clear that logb(b

x) = x, sincex is the number you must raise b to in order to get out bx.

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The real point of statements (1) and (2) above is that you can use the logarithm to“cancel” and exponential function and vice versa.

Example 22.4. The number e = 2.718281 . . . is a special number for logarithms, al-though I won’t be able to tell you why in this course (at least not in a way that youcan understand). It is so important for logarithms that we use the special notation ln(x)(the “natural log” of x) as shorthand for loge(x). The number e has an infinite decimal

expansion, just like π or√

2. The number 3−√

563

shares its first seven digits in common

with e, and so does 6√π4 + π5.

For doing computations with ln, we really need to just rely on the cancellation laws forlogarithms. So, for example ln(e2x) = 2x. Meanwhile, e2 ln(4x) = (eln(4x))2 = (4x)2 = 16x2.The cancellation laws also work for regular logarithms as well. So, for example:

log10(1, 000, 000) = log10(106) = 6

100log10(5) = (102)log10(5) = 102 log10(5) = (10log10(5))2 = (5)2 = 25

Finally, it is also worth pointing out that when we write log(x) without a base, wereally mean log10(x). I’m going to start doing this without further ado.

Homework 22.3 Use the cancellation laws of logarithms and exponents to compute orsimplify the following:

(a) log(107)

(b)(

110

)log(35)(c) y ln(e4x)(d) 1000log(y)

(e) (√

10)log(4)

(f) 8log2(3)

For more practice try problems 53–72 of Chapter 9.3.

22.3. The graph of a logarithm. What the previous subsection tells us is that logb(x)is the inverse function of bx, in math jargon. This means that the graph of logb(x) canbe obtained by rotating the graph of bx 180 degrees in space around the line y = x inthe x-y plane. As you can see (or will once I draw this on the board) the graph of everylogarithmic function has the following properties:

• It is not defined for x ≤ 0, it’s domain consists only of the set of postive x.• It has an asymptote at the y-axis, it shoots down to −∞.• It always has a unique x intercept at x = 1.


• It always passes through the point (b, 1).• It goes to infinity as you go right, but very slowly.

Homework 22.4: Draw a rough sketch of the following. Mark out four distinct pointson each explicitly.

(a) f(x) = log2(x)(b) f(x) = ln(x)

23. Logarithms II

This section of the notes corresponds to Section 9.4 of the book. The purpose is toexplain the following 4 fundamental laws of logarithms, and make you comfortable usingthem.

(1) logb(xy) = logb(x) + logb(y)(2) logb(x/y) = logb(x)− logb(y)(3) logb(x

r) = r · logb(x)(4) loga(b) · logb(c) = loga(c)

23.1. The addition and subtraction laws. First let’s see why the first two logarithmlaws are true. All positive numbers x and y can be rewritten as powers of b, so let’ssuppose that x = br and y = bs. Then for the addition law, we find

logb(xy) = logb(brbs) = logb(b

r+s) = r + s = logb(br) + logb(b

s) = logb(x) + logb(y)

In the first equality, we made the substitutions x = br and y = bs. In the second equalitywe used the exponent laws. In the third equality we used the inverse law for logarithms(covered at the end of the previous class). In the fourth we used the inverse law again,in reverse. Then in the fifth equality we substituted x and y back in for br and bs. Thederivation of the second law is similar, I’ll give it here without detailed explanation:

logb(x/y) = logb(br/bs) = logb(b

r−s) = r − s = logb(br)− logb(b

s) = logb(x)− logb(y).

There are two ways in which the addition and subtraction laws are used: to expand andto condense.

Example 23.1. Let’s expand the expression log2

(4xz

). This works like so:

log2

(4x

z

)= log2(4x)− log2(z) = log2(4) + log2(x)− log2(z).

Since log2(4) = 2, the expression simplifies down to give:

log2

(4x

z

)= 2 + log2(x)− log2(z).

Example 23.2. Now let’s do ln(x2

xy

)(remember ln(z) = loge(z)). I’ll do this two ways.

The first, easy way is just to simplify the fraction inside the parentheses to get

ln

(x2

xy

)= ln

(x

y

)= ln(x)− ln(y).


However, even if we didn’t see the simplification at the beginning, the log rules alonewould have gotten us the same answer:

ln

(x2

xy

)= ln(x2)−ln(xy) = ln(x·x)−(ln(x)+ln(y)) = ln(x)+ln(x)−ln(x)−ln(y) = ln(x)−ln(y).

As you can see, the parentheses were important to remember when applying the additionrule to − ln(xy) in the second equation.

Homework 23.1: Expand and simplify the following logarithmic expressions as muchas possible.

(a) log4(4x)(b) log2(7/8)(c) log5(50)(d) ln(e2/5)

(e) log(

xy1,000

)(remember that log(x) really means log10(x).)

(f) log2(2−3)

For more practice, try problems 1-36 of Chapter 9.4. However, keep in mind that someof them will require you to use the third rule of logarithms, which we will cover shortly.

Log rules (1) and (2) can be used in the opposite direction as well, to condense expres-sions down to a single one.

Example 23.3. Let’s condense log2(x) − log2(7) + log2(3) down to a single logarithm.This goes like so:

log2(x)− log2(7) + log2(3) = log2

(x7

)+ log2(3) = log2

(x7· 3)

= log2

(3x

7

).

WARNING: One common error when performing this kind of calculation is to try to startwith the following move:

log2(x)− log2(7) + log2(3) = log2(x)− log2(7 · 3).

This leads to the incorrect answer log2

(x21

). Why? Because you have applied the addition

rule to − log2(7) + log2(3), but this is wrong because the addition rule does not allow anegative sign in front of either logarithm. Rather, the correct thing to do here is to usethe subtraction rule, since

− log2(7) + log2(3) = log2(3)− log2(7) = log2

(3

7

).

The lesson here is that when you apply the addition and subtraction rules, you shouldnot focus on the order in which the various terms appear. Instead, just remember thatthe negative terms always end up in the denominator and the positive logs always end upin the numerator, no matter what order they appear in.

Homework 23.2: Condense the following expressions to a single logarithm with nonegative sign in front.

(a) log(x)− log(13)(b) log2(x)− log2(y)− log2(z) + log2(7)


(c) log3(x2 + 1)− log3(x)

(d) 2 log(7)− 2 log(5)(e) − log8(x)− log8(y)

For more practice, try problems 37-60 of Section 9.4, although again be aware thatsome of them won’t be solvable without rule (3) for logarithms.

23.2. The “pull down” rule for logarithms. The third rule for logarithms, logb(xr) =

r logb(x) can be shown to be true as follows. Suppose x = bs, then

logb(xr) = logb((b

s)r) = logb(brs) = rs = r logb(b

s) = r logb(x).

The justification of each inequality is very similar to the justification used in each step ofthe proof of the addition rule in the previous section. I urge you to try and figure out foryourself why each step must be true, it is a good test of your understanding.

Notice that the pull down rule is consistent with the addition rule. For example, usingthe addition rule, we get

logb(x3) = logb(x · x · x) = logb(x) + logb(x) + logb(x) = 3 logb(x)

which could have been derived in one step using the “pull-down” rule. It’s a little moreinteresting to see how nicely the rule works for negative exponents. For example, usingthe “pull down” rule we get

logb(x−2) = −2 logb(x).

Using just the addition and subtraction rules we get the same thing, but with more work:

logb(x−2) = logb

(1

x2

)= logb(1)− logb(x

2) = 0− logb(x)− logb(x) = −2 logb(x)

where here we used the fact that logb(1) = 0 for any base b.

The pull down rule is easy to use in either direction, so long as you remember two bigrules. First you must remember to switch from radical notation to exponent notation.

Example 23.4. Let’s expand ln(23√xy) as much as possible. We’ll start by using theaddition rule to get

ln(23) + ln(√xy)

. The pull-down rule is easy to use on the first term. For the second term, we will switchthe radical to exponent notation first, then we can use the pull-down rule on it as well,like so:

ln(23) + ln((xy)1/2) = 3 ln(2) +1

2ln(xy).

From there we can use the addition rule on ln(xy), to get

3 ln(2) +1

2ln(x) +

1

2ln(y).

The second big rule is that you can only pull an exponent down if it is on top ofeverything inside the logarithm’s brackets.


Example 23.5. You cannot apply the exponent rule to log(2x3) to get 3 log(2x), becausethe 2 was not raised to the third power. On the other hand, you can apply the exponentrule to log(8x3) = log(23x3), but not one at a time like this log(23x3) = 3 log(2x3) =9 log(2x). Instead you have to group the stuff under the exponents into one group likethis:

log(8x3) = log(23x3) = log((2x)3) = 3 log(2x).

You are now ready to fully expand or fully condense most logarithmic expressions.Homework 23.3 Full expand and simplify the following expressions.

(a) logb

(x2yz2

)(b) ln(

√ex)

(c) log

(3√100x2

30 4√y3

)(d) log3

(5

√x3

5y2

)Homework 23.4 Fully condense and simplify the following expressions.

(a) 12

ln(4) + 2 ln(x2)(b) 5 ln(x)− 2 ln(y)(c) 1

2(log2(x)− 2 log2(y)) + log2(y)

24. Logarithms, III

In this section of the notes we finish up Section 9.4 of the book and cover 9.5 of thebook. This will complete the lessons about logarithms.

24.1. The change of base formula. Most scientific calculators can only calculate ln(x)and log10(x) explicitly. However, using the change of base formula, you can calculate thelogarithm at any base.

Change of Base Formula: For any three positive numbers a, b, and c, the followingequation is true:

(24.1) loga(b) logb(c) = loga(c)

Or, equivalently:

(24.2) logb(c) =loga(c)

loga(b)

Proof. To see that the first equation is true, we raise a to the power of each side to getthe new, equivalent equation:

aloga(b) logb(c) = aloga(c)

Using the cancellation law, we see that the right side immediately becomes c. Meanwhilethe left side also reduced to c in the following sequence of cancellations aloga(b) logb(c) =(aloga(b)

)logb(c) = blogb(c) = c.The second version of the equation comes from dividing both sides by loga(b).

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Example 24.1. Let’s use the change of base formula and a calculator to compute log2(7).In Equation , set 2 = b and 7 = c. Since my calculator computes logs in base 10, I’ll seta = 10. The change of base formula then gives me

log2(7) =log10(7)

log10(2).

Now I have an expression that I can actually plug into a calculator, I get an answer ofapproximately 2.807. To check that this answer makes sense, I plug in 22.807 = 6.998,which is close enough to say that my approximation is good.

Homework 24.1: Use the change of base formula and your calculator to compute thefollowing logarithms.

(a) log2(8) (the correct answer is of course 3, but I want you to use your calculator tofind it.)

(b) log7(π)(c) log3(2)(d) log3(8) (note: this should be approximately 3 times larger than your answer to c!)(e) log5(1/3)

24.2. Solving exponential equations. An exponential equation is just an equationwhich involves an exponential function ax. The basic strategy for solving such equationsis to lower all exponents using logarithms.

Example 24.2. Let us solve the equation 2x = 13. If we take the (base 10) logarithm ofboth sides, we get a new, equivalent equation

log(2x) = log(13).

Using the “pull down” rule, we then get

x · log(2) = log(13).

If we now divide both sides of this equation by log(2) we get

x =log(13)

log(2)≈ 3.7

We can verify that this approximation is good by plugging 3.7 back in for x in the originalequation to get 23.7 = 12.996 ≈ 13.

An alternative and equally valid approach would be to start by taking log2 of both sidesof the original equation 2x = 13, to get the new equation

log2(2x) = log2(13).

This has the advantage that we can immediately reduce the left side to x using the logcancellation law occurring in Proposition 22.3 above, so that we get

x = log2(13).

Unfortunately, log2(13) cannot be plugged directly into the calculator, so we will have toconvert log2(13) to base e or 10 to finish the calculation anyway, which will give us theanswer I obtained above.


Example 24.3. Now let’s do a real problem. Let’s solve the equation 3 · ex = 5x+1.Because there is an e appearing on the left side, I am going to apply the natural logarithmln = loge to both sides. This gives me

ln(3 · ex) = ln(5x+1).

It is tempting to use the pull down rule immediately on the left side of this equation, butit is not valid to do so here. Instead we must first expand the left side using the additionrule to get

ln(3) + ln(ex) = ln(5x+1).

We can use the cancellation law of Proposition 22.3 to change ln(ex) = loge(ex) into x,

and we can use the pull down rule on ln(5x+1) to get (x + 1) ln(5). This changes ourequation into

ln(3) + x = (x+ 1) · ln(5).

At this point, all the exponents have been pulled down, so now we are just trying to solvefor x using old fashioned techniques. For this purpose, remember that ln(3) and ln(5) arejust numbers, in fact if you wanted you could plug those numbers into your calculatorright now and finish the problem without any logarithms at all. However, I’m going toavoid computing them until the end.

(24.3)

ln(3) + x = (x+ 1) · ln(5)ln(3) + x = x · ln(5) + ln(5)

ln(3)− ln(5) = x · ln(5)− xln(3)− ln(5) = x(ln(5)− 1)

ln(3)−ln(5)ln(5)−1 = x

Plugging this last expression into our calculator, we conclude that x ≈ −0.838. Sincewe’ve done a lot of steps here, it is a good idea to verify that this approximation is correctby plugging it back into each side of the original equation to get 3 · e−0.838 ≈ 1.3 and50.162 ≈ 1.3, which coincide.

Homework 24.2: Solve the following equations for x. Use your calculator to give adecimal answer.

(a) 10x = 13(b) 3x = 21(c) 5ex = 14(d) 4− (1.4)x = 0(e) 2x = 3x−1

(f) 4ex = 7x+2

For more practice, try problems 19-40 of Section 9.5 in the book.

24.3. Logarithmic equations. A logarithmic equation is one which contains logarithmsin it. The basic idea for solving it is to use exponentiation to cancel out the logarithms,similar to what we did in the proof of the change of base formula above.

Example 24.4. To solve log3(x) = −4, we take 3 and raise it to the power of each sideof the equation to get a new, equivalent equation

3log3(x) = 3−4.


We can now use the cancellation law of Proposition 22.3 to reduce the left side to just x.We get out

x = 3−4 =1

34=

1

81.

More interesting logarithmic equations require us to condense the logarithms first beforeusing exponentiation to cancel them out.

Example 24.5. Let’s solve 2 log3(x+1)− log3(x+2) = −1. We could use exponentiationright away if we like, but things will generally be cleaner if we use the log rules first tocondense the left side to a single logarithmic expression. For this, note that

2 log3(x+ 1)− log3(x+ 2) = log3((x+ 1)2)− log3(x+ 2) = log3

((x+ 1)2

(x+ 2)

).

Hence our equation becomes

log3

(x2 + 2x+ 1

x+ 2

)= −1.

We now raise 3 by each side and apply cancellation on the left to get out the equivalentequation

x2 + 2x+ 1

x+ 2= 3−1.

This is now a standard rational equation which we know how to solve using older tech-niques. Multiplying both sides by 3(x+ 2) gives

3x2 + 6x+ 3 = x+ 2.

We now subtract x+ 2 from both sides to get a quadratic equation in standard form

3x2 + 5x+ 1 = 0.

The quadratic formula tells us that this has solutions

x =−5

6±√

13

6.

The solution −56−√136

is inadmissable because it makes one of the logarithms in theoriginal equation undefined when you plug it it, so we are left with the single solution

x =−5 +

√13

6≈ −.2324

Homework 24.3: Solve for x (you do not need to use your calculator to give a decimalapproximation on these ones).

(a) log16(x) = −14

(b) 5 ln(2x) = 20(c) log5(x) + log5(4x− 1) = 1(d) log(x+ 4)− log(2) = log(5x+ 1)(e) log2(x− 1)− log2(x+ 3) = log2(1/x)


25. A new kind of number

Today we go back to Section 7.7 of the book to cover imaginary numbers.


25.1. The number i and its basic arithmetic. Up to now, I’ve always maintained thatif a is a positive number, then

√−a is undefined. And it is true that we cannot define any

real number to be the square root of a negative number, since real numbers only squareto positive numbers. However, if we are willing to invent a new kind of number, then wecan make

√−a defined after all.

Definition 25.1. Let i be a new number which, by definition, satisfies the equationi2 = −1. Equivalently, define i =

√−1.

This number i is called an imaginary number, to distinguish it from the “real” numberswe are used to, and which correspond to magnitudes in space. Since we know nothingabout this new number i except that i2 = −1, we treat it like a variable x in our calcula-tions until the end, when we simplify using the rule i2 = −1.

Example 25.2. Let’s compute (2 + i) · (3 − i). We start by foiling out the expressions,treating i like a variable x, to get

6 + i+ i2.

We can now use the fact that i2 = −1 by definition to simplify the expression downfurther to 6 + i− 1. We end by concluding

(2 + i) · (3− i) = 5 + i.

In fact, any standard arithmetic expression involving the new number i can eventuallybe reduced in this way to an expression of the form a+bi, where a and b are real numbers.

Homework 25.1: Reduce each of the following algebraic expressions to one of the forma+ bi

(a) (3 + 2i)− (4− i)(b) (5 + 6i) · (1− 2i)(c) (3− 2i)2

(d) (5 + i√

3) · (5− i√

3)

For more practice try exercises 17-54 of Section 7.7.

So it turns out we can do arithmetic with these new imaginary numbers. They alsoallow us to work with negative radicands.

Example 25.3. Let us simplify√−18 as much as possible. Just like with a positive

number, we factor −18 as much as possible, then apply the radical to certain piecesseparately √

−18 =√

(−1)32 · 2 =√−1 ·√

32 ·√

2 = i3√

2.

In other words, the one new fact you need to use when dealing with negative radicands isthat −1 turns into i when you pull it out of the radical. So

√−2 = i

√2,√−4 = i

√4 = 2i,

etc.Homework 25.2 Simplify the following expressions:

(a)√−100

(b) 7 +√−4

(c)√−7 ·√−3

(d) i√−6 ·√−3 ·√−2


For more practice, try problems 1-14 and 55-62 of Section 7.7

25.2. Complex numbers and some next-level arithmetic. A number of the forma + bi, where a and b are real numbers, is called a complex number. In the previoussubsection we saw that it is possible to add any two complex numbers to together toget another complex number, and it is also possible to multiply two complex numberstogether to get out another complex number. But we did not figure out how to dividecomplex numbers? Is it possible? Yes it is.

Example 25.4. Let’s compute 2−3i2i

first. Our problem is with the i in the denominator,once we get rid of it, we can rewrite this expression in the form a + bi and be done. Toget it, we multiply top and bottom by i to get

2− 3i

2i· ii

=2i− 3i2

2i2=

3 + 2i

−2= −3

2− i

So there’s an example of how division works with an imaginary number of the form bi.What about dividing by a number of the full form a+ bi?

Example 25.5. Now let’s do this one: 4+i1−3i . We use an old trick for rationalizing denom-

inators, multiplying top and bottom by the conjugate.

4 + i

1− 3i· 1 + 3i

1 + 3i=

1 + 13i

10=

1

10+

13

10i

Homework 25.3: Compute the following complex numbers so that they have the forma+ bi

(a) 1i

(b) 1+i7i

(c) 7−4i1+2i

(d) 2+i7−i

For more practice try Problems 63-84 of Section 7.7.

At this point you might be wondering what all of this multiplication and division reallycorresponds to. I’ll tell you in class if you choose to attend.

E-mail address: [email protected] address: [email protected]

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