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FLC Ch 5
Page 1 of 13
Math 100 Elementary Algebra Sec 5.1: The Greatest Common Factor and Factor By Grouping (FBG)
Ex 1 Factor. (Check work by multiplying.) a) b)
21π4 β 14π3 + 35π2 96π₯2π¦2 β 144π₯3π¦ + 48π₯π¦ c) d)
6(3π + π) β π§(3π + π) 3π(ππ β 3π) β 12(ππ β 3π) β 6π(ππ β 3π) e) f) Factor out 1/3.
7π2(π + 4π) + π + 4π 2
3π₯2(2π₯ β 1) β
4
3π₯(2π₯ β 1) + 3(2π₯ β 1)
g) h) 20π3π3 β 18π3π4 + 22π4π4 6π₯ β 3π₯π¦ + 9π¦ Ex 2 PP Find the area of the shaded region in factored form. π is the radius of the larger circle and π is the radius of the smaller circle. Ans: π (πΉπ β ππ)
Recall: In the product ππ, π and π are factors. Defn In an expression, any factor that is common to each term is called a common factor. The largest of all common factors is called the greatest common factor (GCF). Remark: The answer upon factoring is always a ___________________.
When factoring, we ALWAYS start with the ___________________________ (unless itβs 1).
FLC Ch 5
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Practice Problems Factor. 1) 6π₯π¦ β 15π§ + 21 2) 20π₯2 β 32π₯π¦ + 12π₯ 3) 7(4π₯ β 5) β π(4π₯ β 5)
4) 2π₯(8π¦ + 3π§) β 5π¦(8π¦ + 3π§) 5) 4π₯3(π₯ β 1) β (π₯ β 1) Ex 3 Factor. a) π₯π¦ β 4π₯ + 3π¦ β 12 b) 10ππ β 3π + 5ππ β 6π c) π₯2 β 2π₯ β π₯π¦ + 2π¦ d) 15π3 β 25π2π β 18ππ2 + 30π3 e) ππ₯ + ππ₯ + ππ₯ + ππ¦ + ππ¦ + ππ¦ Practice Problems Factor. 1) 18 + 3π₯ β 6π¦ β π₯π¦ 2) 15π₯ β 9π₯π + 20π€ β 12ππ€ 3) π₯3 β 5π₯2 β 3π₯ + 15 4) 7π + 21π + 2ππ + 6π2 5) 30π3 + 12π2π β 25ππ2 β 10π3 Good Start?: (ππ β πππ) β (ππ + ππ)
Sec 5.2: Factoring Trinomials of the Form ππ + ππ + π (and 5.7)
We will βdissectβ the FOIL method to factor trinomials. Consider different combinations of (π₯ 2)(π₯ 5). Observe numbers and signs. Ex 4 Factor. a) π₯2 + 12π₯ + 32 b) π¦2 β 16π¦ + 60 c) π¦2 + 11π¦ β 60
Factoring Trinomials of the Form ππ + ππ + π The factorization will have the form (π₯ + π)(π₯ + π) where ππ = π and π + π = π.
Use FBG method when factoring a polynomial with 4 (or more) terms.
FLC Ch 5
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d) π¦2 β 11π¦ β 60 e) 2π₯2 + 2π₯ + 12 f) 6π₯2 + 24π₯ + 18 How can we adjust?
g) 3π₯2π¦ β 6π₯π¦ + 21π₯π§ h) π₯8 β 2π₯4 β 15 i) π₯2 β2
3π₯ +
1
9
PP PP
j) π₯2 + 0.8π₯ + 0.15 k) If π₯ β 4 is a factor of π₯2 + ππ₯ β 20, what is the value of π?
Sec 5.7: Solving Quadratic Equations by Factoring
Exs Solve for π₯.
(π₯ β 3)(π₯ + 2) = 0
1
2π₯(2π₯ β 1)(3π₯ + 4) = 0
Facts About Signs
Each represents some positive number.
π₯2 + π₯ + will factor as (π₯ +)(π₯ +) π₯2 β π₯ + will factor as (π₯ β)(π₯ β) π₯2 + π₯ β will factor as (π₯ +)(π₯ β) π₯2 β π₯ β will factor as (π₯ +)(π₯ β)
Steps to Solve a Quadratic Equation by Factoring β Use ZFP
1) Make sure the equation is set to 0. 2) Factor, if possible, the quadratic expression. 3) Set each factor containing a variable equal to 0. 4) Solve the resulting equations to find each root. 5) Check each root.
Defn A quadratic equation is an equation of the form ππ₯2 + ππ₯ + π = 0, where π, π, and π are real numbers and π β 0. ππ₯2 + ππ₯ + π = 0 is the standard form of a quadratic equation.
Zero Factor Property
If π β π = 0, then π = 0 or π = 0 (or both)
Last Sign of any Poly um ame signs ifference ifferent signs
S
D
FLC Ch 5
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Ex 5 Solve. a) π₯2 β 6π₯ β 7 = 0 b) 2π₯2 β 12π₯ = 54 c) π₯(π₯ + 1) = 110
ππ(π β π) = ππ
Ex 6 Find the area of the shaded region of the figure below in factored form. The dimensions of the smaller rectangle are π₯ Γ (π₯ + 2). Practice Problems Factor. 1) π₯2π¦ + 14π₯π¦ + 48π¦ 2) 2π₯2 β 12π₯ β 54
Sec 5.3: Factoring Trinomials of the Form πππ + ππ + π AND
Sec 5.4: The Difference of Two Squares and Perfect Square Trinomials Ex 7 Factor using the trial-and-error method.
a) 3π₯2 + 11π₯ + 10 b) 12π₯2 + 5π₯ β 3 c) 4π₯2 β 7π₯ β 15 What multiplies to 10 and adds to 11?
Ex 8 ____________ 4π₯2 β 7π₯ β 15 = 0 Ex 9 Factor. a) 3π₯2 β 13π₯ β 10 b) 12π₯2 + 7π₯ β 12
12
10
FLC Ch 5
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c) 36π₯2 β 1 d) 25π2 β 64π2 e) 25π₯2 + 10π₯ + 1 ππππ + ππ β π
f) 25π₯2 β 10π₯ + 1 g) 9 β 100π¦2 h) 36π₯2 + 1 i) If 2π₯ β 5 is a factor of 6π₯2 + ππ₯ + 10, what is the value of π? Ex 10 #68 At the beginning of every football game, the referee flips a coin to see who will kick off. The equation that gives the height (in feet) of the coin tossed in the air is β = 6 + 29π‘ β 16π‘2. a) Factor the equation. b) Use the factored form of the equation to find the height of the quarter after 0 seconds, 1 second, and 2 seconds.
Factoring the Difference of Two Squares
π2 β π2 = (π + π)(π β π) Note: π2 + π2 is ________.
Perfect Square Trinomials
π2 + 2ππ + π2 = (π + π)2 π2 β 2ππ + π2 = (π β π)2
FLC Ch 5
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Ex 11 Factor or solve. 12π₯2 + π₯ β 6 12π₯2 = βπ₯ + 6 π₯2 β π₯ + 5 = 0 Can this appear on exam 2?
In-Class Problems/Quiz: How Am I Doing?
1) 3π2 β 10π β 8 2) 10π₯2 + π₯ β 2 3) 3π₯2 β 23π₯ + 14 4) 4π₯2 β 11π₯ β 3 5) 12π₯2 β 24π₯ + 9 6) 20π₯2 β 38π₯ + 12 7) 6π₯2 + 17π₯π¦ + 12π¦2 8) 14π₯3 β 20π₯2 β 16π₯ 9) 24π₯2 β 98π₯ β 45 (Quiz EC)
Ex 12 Factor. a) 36π₯2 + 60π₯π¦ + 25π¦2 b) 121π¦2 β 49 c) 50π2 β 160ππ + 128π2 d) 5π₯2 + 40π₯ + 80 e) 2π₯2 β 32π₯ + 110 f) π¦2π§ β 12π¦π§ + 36π§ g) π₯2 + 16 h) π₯2 β 16 i) 2π₯4 β 32
FLC Ch 5
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Practice Problems Factor. 16π₯2 β 36π¦2 49π₯2 β 28π₯ + 4 100π₯2 β 9 18π¦2 β 50π₯2 25π₯2 + 20π₯ + 4 25π₯2 β 20π₯ + 4 49π₯2 β 28π₯π¦ + 4π¦2 3π₯2 β 75 72π₯2 β 192π₯ + 128 144π₯2 Β± ________ + 81π¦2
Sec 5.6: Factoring: A General Review AND Sec 5.7: Solving Quadratic Equations by Factoring Refer to βFactoring Polynomials Guideβ. Indicate number of terms for each type.
Factor completely or solve for the roots of each quadratic equation. If the polynomial is not
factorable, you must state that itβs prime. Check answers! How? What will each answer look like?
14) 3ππ₯ + 9ππ₯ β 12ππ€ β 36ππ€ 15) π₯3 + 2π₯2π¦ β 15π₯π¦2 16) 8 + 7π₯ β π₯2 Do DO Do 17) 4π₯2 + 2π₯ = 0 18) π₯2 + π₯ β 42 19) 7π₯2 β 252 Do PP Do 20) π₯2 + 36 21) π₯2 β 7π₯ β 14 22) 4π₯2 β 4π₯ β 80 = 0 Do Do Do 23) 8π₯3 β 22π₯2 + 5π₯ = 0 24) 100π₯2 + 25 25) 10π₯2 + π₯ β 2 Do Do Do
FLC Ch 5
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26) 25π₯2 + 16π¦2 27) 64π₯2 + 48π₯ = β9 28) π₯3 β 5π₯2 β 4π₯ + 20 Do Do
29) 2π₯2 β 10π₯ β 14 30) 3π₯2 β 33π₯ + 54 31) 4π₯4 β 11π₯2 β 3 Do 32) 18π₯2 β 69π₯ + 60 33) 2π₯2 + π₯ + 6 34) 12π₯2 + 11π₯π¦ β 5π¦2 Do Do 35) 4π₯2 β 13π₯ β 12 36) π₯2 + 7π₯ + 1 37) (π₯ + 3π¦)2 β 16 Do Do ππ β (π + ππ)π
FLC Ch 5
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38) 8π₯π€ + 9π₯2 + 35π₯π¦2 + 28π¦2π€ + π₯2 39) 10π₯2 + 5π₯π¦ β 20 40) 18 β 2π₯2 Start Do Ans: (ππ + ππ)(ππ + πππ) 41) 25π₯2 = 36 42) (π₯ β 3)4 + 4(π₯ β 3)2 43) (3π₯ β 2)3 β 3π₯ + 2 Ex 44 Solve and check. a) Do b) Do
(2π₯ β 3)(π₯ β 1) = 3 (π₯ β 5)(π₯ + 4) = 2(π₯ β 5) c) Do d) Do e) PP
π₯2 + 5π₯
6= 4 (3π₯ β 4)(5π₯ + 1)(2π₯ β 7) = 0 (1119π₯ β 1)(777π₯ + 19) = 0
Ans: π =π
ππππ, β
ππ
πππ
FLC Ch 5
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f) g) h) π₯(12 β π₯) = 32 152 = (π₯ + 3)2 + π₯2 4π₯3 + 12π₯2 β 9π₯ β 27 = 0 Ex 45 PP Grade the solution. 81π₯2 β 16 = (9π₯ + 4)(9π₯ β 4) = (3π₯ + 2)(3π₯ β 2) Ex 46 Fill in the boxes to create a perfect square trinomial. 64π₯2 Β± 81π¦2 Ex 47 Consider (29π₯ + 7)(29π₯ β 14) = 0, (29π₯ + 7)(29π₯ β 14) = 1, and (29π₯ + 7)(29π₯ β 14) = π₯.
Sec 5.8: Applications of Quadratic Equations Ex 48 (#6) The product of two consecutive odd integers is 1 less than 4 times their sum. Find the two integers. Define variable and set up. PP-solve. Ans: 7 and 9 OR -1 and 1
FLC Ch 5
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Ex 49 (#12) One number is 2 more than twice another. Their product is 2 more than twice their sum. Find the numbers. Ex 50 (#14) The length of a rectangle is 3 more than twice the width. The area is 44 square inches. Find the dimensions. Ex 51 (#18) The hypotenuse of a right triangle is 15 inches. One of the legs is 3 inches more than the other. Find the lengths of the two legs.
Pythagorean Theorem In any right triangle, if π is the length of the hypotenuse and π and π are the lengths of the two legs, then π2 + π2 = π2.
π
π π
FLC Ch 5
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Ex 52 (#34) A rocket is fired vertically into the air with a speed of 240 feet per second. Its height at time π‘ seconds is given by β(π‘) = β16π‘2 + 240π‘. At what time(s) will the rocket be the following number of feet above the ground? a) PP 704 feet b) 896 feet c) Why do parts a and b have two answers? d) How long will the rocket be in the air? e) When the equation for part d is solved, one of the answers is π‘ = 0. What does this represent? Ex 53 (#26) A company manufactures flash drives for home computers. It knows from experience that the number of drives it can sell each day, π₯, is related to the price π per drive by the equation π₯ = 800 β 100π. At what price should the company sell the flash drives if it wants the daily revenue to be $1200? The equation for revenue is π = π₯π.
FLC Ch 5
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Revisit example 1a
Multiplicity Ex 54 PP You are standing on the edge of a cliff near Acapulco, overlooking the ocean. The place where you stand is 180 meters from the ocean. You drop a pebble into the water. (Dropping the pebble implies that there is no initial velocity, so π£ = 0.) How many seconds will it take to hit the water? How
far has the pebble dropped after 3 seconds? Use the formula πΊ = βπππ + ππ + π, where π = the height of the object π£ = the upward velocity in meters/second π‘ = the time of flight in seconds β = the height above level ground from which the object is thrown
Discriminant βProblems from Factoring Assignment (Due:________________)