Plowing Through Sec. 2.4bPlowing Through Sec. 2.4bwith Two New Topics:with Two New Topics:

Synthetic DivisionSynthetic Division

Rational Zeros TheoremRational Zeros Theorem

Synthetic DivisionSynthetic Division

Synthetic DivisionSynthetic Division is a shortcut method for the is a shortcut method for thedivision of a polynomial by a linear divisor, division of a polynomial by a linear divisor, xx – – kk..

Notes:Notes:

This technique works This technique works only only when dividing by awhen dividing by alinear polynomial…linear polynomial…

It is essentially a “collapsed” version of the longIt is essentially a “collapsed” version of the longdivision we practiced last class…division we practiced last class…

Synthetic Division – Examples:Synthetic Division – Examples:3 22 3 5 12x x x

3x Evaluate the quotient:Evaluate the quotient:

33 22 ––33 ––55 ––1212

22 33 44 00

66 99 1212

Coefficients of dividend:Coefficients of dividend:Zero ofZero ofdivisor:divisor:

RemaindeRemainderr

QuotientQuotient3 22 3 5 12x x x

3x 22 3 4x x

Synthetic Division – Examples:Synthetic Division – Examples:4 22 3 3x x x 2x Divide by and write a

summary statement in fraction form.

––22 11 00 ––22 33

4 23 22 3 3 12 2 1

2 2

x x xx x x

x x

––33

11 ––22 22 ––11 ––11

––22 44 ––44 22

Verify Graphically?Verify Graphically?

Rational Zeros TheoremRational Zeros Theorem

Real zeros of polynomial functions are either Real zeros of polynomial functions are either rationalrationalzeroszeros or or irrational zerosirrational zeros. Examples:. Examples:

24 9f x x 2 3 2 3x x The function has The function has rational zerosrational zeros –3/2 and 3/2 –3/2 and 3/2

2 2f x x 2 2x x The function has The function has irrational zerosirrational zeros – 2 and 2 – 2 and 2

Rational Zeros TheoremRational Zeros TheoremSuppose f is a polynomial function of degree n > 1 of the form

11 0

n nn nf x a x a x a

with every coefficient an integer and . If x = p /q isa rational zero of f, where p and q have no common integerfactors other than 1, then

0 0a

• p is an integer factor of the constant coefficient , and0a

• q is an integer factor of the leading coefficient .na

RZT – Examples:RZT – Examples:Find the rational zeros of 3 23 1f x x x The leading and constant coefficients are both 1!!!

The only possible rational zeros are 1 and –1…check them out:

3 21 1 3 1 1 3 0f 3 21 1 3 1 1 1 0f

So So ff has no rational zeros!!! has no rational zeros!!!(verify graphically?)(verify graphically?)

RZT – Examples:RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x

Potential Rational Zeros:

1, 2 Factors of –2

Factors of 3 1, 3 1 2

1, 2, ,3 3

Graph the function to narrow the search…

Good candidates: 1, – 2, possibly –1/3 or –2/3

Begin checking these zeros, using synthetic division…

RZT – Examples:RZT – Examples:Find the rational zeros of 3 23 4 5 2f x x x x

1 3 4 –5 –2

3 7 2

3 7 2 0

Because the remainder is zero,x – 1 is a factor of f(x)!!!

23 7 2 1f x x x x Now, factor the remaining quadratic…

3 1 2 1f x x x x

The rational zeros are 1, –1/3, and –2The rational zeros are 1, –1/3, and –2

RZT – Examples:RZT – Examples:Find the polynomial function with leading coefficient 2 that hasdegree 3, with –1, 3, and –5 as zeros.

First, write the polynomial in factored form:

2 1 3 5f x x x x

Then expand into standard form:

22 2 3 5f x x x x

3 2 22 5 2 10 3 15x x x x x 3 22 6 26 30x x x

RZT – Examples:RZT – Examples:Using only algebraic methods, find the cubic function with thegiven table of values. Check with a calculator graph.

x

2 1 5f x k x x x

–2 –1 1 5

f(x) 0 24 0 0

(x + 2), (x – 1), and (x – 5)must be factors…

But we also have : 1 24f 1 2 1 1 1 5 24k 2k

2 2 1 5f x x x x

22 2 5x x x 3 22 8 14 20x x x

A New Use for SyntheticA New Use for SyntheticDivision in Sec. 2.4:Division in Sec. 2.4:

Upper andUpper andLower BoundsLower Bounds

What are they???

A number k is an upper bound for the real zeros off if f (x) is never zero when x is greater than k.

A number k is a lower bound for the real zeros off if f (x) is never zero when x is less than k.

What are they???

Let’s see them graphically:

y f x

c

d

cc is a is a lower bound lower bound and and dd is an is an upper upper boundbound for the real for the realzeros of zeros of f f

Upper and Lower Bound Tests for Real Zeros

Let f be a polynomial function of degree n > 1 with a positiveleading coefficient. Suppose f (x) is divided by x – k usingsynthetic division.

• If k > 0 and every number in the last line is nonnegative (positive or zero), then k is an upper bound for the real zeros of f.

• If k < 0 and the numbers in the last line are alternately nonnegative and nonpositive, then k is a lower bound for the real zeros of f.

Cool Practice Problems!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function must lie inthe interval [–2, 5].

4 3 22 7 8 14 8f x x x x x The function has a positive leading coefficient, so we employour new test with –2 and 5:

5 2 –7 –8 14 8

10 15 35 245

2 3 7 49 253

–2 2 –7 –8 14 8

–4 22 –28 28

2 –11 14 –14 36

This last line is all positive!!! This last line hasalternating signs!!!

5 is an upper bound5 is an upper bound – –2 is a lower bound2 is a lower bound

Let’s check these results Let’s check these results graphicallygraphically……

Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.

4 3 22 7 8 14 8f x x x x x From the last example, we know that all of the rational zerosmust lie on the interval [–2, 5].

Next, use the Rational Zero Theorem…potential rational zeros:

Factors of 8

Factors of 2

1, 2, 4, 8

1, 2

11, 2, 4, 8,

2

Look at the graph to find likely candidates: Let’s try 4 and –1/2

Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.

4 3 22 7 8 14 8f x x x x x 4 2 –7 –8 14 8

8 4 –16 –8

2 1 –4 –2 0

3 24 2 4 2f x x x x x –1/2 2 1 –4 –2

–1 0 2

2 0 –4 0

214 2 4

2f x x x x

Cool Practice Problems!!!Cool Practice Problems!!!Find all of the real zeros of the given function.

4 3 22 7 8 14 8f x x x x x

214 2 4

2f x x x x

212 4 2

2f x x x x

12 4 2 2

2f x x x x x

The zeros of f are the rational numbers 4 and –1/2 and theirrational numbers are – 2 and 2

Cool Practice Problems!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function lie in theinterval [0, 1], and find them. 5 210 3 6f x x x x Check our potential bounds:

0 10 0 0 –3 1

0 0 0 0

10 0 0 –3 1

0 is a lower bound!!!0 is a lower bound!!!

–6

0

–6

1 10 0 0 –3 1

10 10 10 7

10 10 10 7 8

1 is an upper bound!!!1 is an upper bound!!!

–6

8

2

Cool Practice Problems!!!Cool Practice Problems!!!Prove that all of the real zeros of the given function lie in theinterval [0, 1], and find them. 5 210 3 6f x x x x Possible rational zeros:

1 3 1 2 3 6 1 31, 2, 3, 6, , , , , , , ,

2 2 5 5 5 5 10 10

Check the graph (with 0 < x < 1) to select likely candidates…

The function has no rational zeros on the interval!!!The function has no rational zeros on the interval!!!

Lone Real Zero: 0.951Lone Real Zero: 0.951Are there any zeros???