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3

Polynomialand

Equations

Inequalities

and

Rational

Functions

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

1

3.3

Zeros of Polynomial Functions

• Factor Theorem

• Rational Zeros Theorem

• Number of Zeros

• Conjugate Zeros Theorem

• Finding Zeros of a Polynomial Function

• Descartes’ Rule of Signs

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

2

Factor Theorem

For any polynomial function f(x), x – k

is a factor of the polynomial if and only

if (k) = 0.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

3

DECIDING WHETHER x – k IS A

FACTOR

Determine whether x – 1 is a factor of each polynomial.

Example 1

(a)

Solution By the factor theorem, x – 1 will be a

factor if (1) = 0. Use synthetic division and the

remainder theorem to decide.

Use a zero

coefficient for

the missing

term.

(1) = 7

The remainder is 7 and not 0, so

x – 1 is not a factor of (x).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

4

DECIDING WHETHER x – k IS A

FACTOR OF

Determine whether x – 1 is a factor of each polynomial.

Example 1

(b)

Solution

(1) = 0

Because the remainder is 0, x – 1 is a factor.

Additionally, we can determine from the coefficients in

the bottom row that the other factor is

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

5

DECIDING WHETHER x – k IS A

FACTOR OF

Determine whether x – 1 is a factor of each polynomial.

Example 1

(b)

Solution

(1) = 0

Thus,

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

6

Example 2

FACTORING A POLYNOMIAL

GIVEN A ZERO

Factor

into linear

factors if – 3 is a zero of .

Solution Since – 3 is a zero of ,

x – (– 3) = x + 3 is a factor.

Use synthetic division to

divide (x) by x + 3.

The quotient is 6×2 + x – 1, which is the

factor that accompanies x + 3.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

7

Example 2

FACTORING A POLYNOMIAL

GIVEN A ZERO

Factor the following into linear factors if – 3 is

a zero of .

Solution

Factor 6×2 + x – 1.

These factors are all linear.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

8

Rational Zeros Theorem

If

is a rational number written in

lowest terms, and if is a zero of , a

polynomial function with integer

coefficients, then p is a factor of the

constant term and q is a factor of the

leading coefficient.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

9

Proof of the Rational Zeros Theorem

Multiply by

qn and

subtract

a 0 q n.

Factor

out p.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

10

Proof of the Rational Zeros Theorem

This result shows that – a0qn equals the

product of the two factors p and

(anpn–1 + + a1qn–1).

For this reason, p must be a factor of – a0qn.

Since it was assumed that is written in

lowest terms, p and q have no common

factor other than 1, so p is not a factor of qn.

Thus, p must be a factor of a0. In a similar

way, it can be shown that q is a factor of an.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

11

Example 3

USING THE RATIONAL ZEROS

THEOREM

Consider the polynomial function.

(a) List all possible rational zeros.

Solution For a rational number to be a

zero, p must be a factor of a0 = 2 and q

must be a factor of a4 = 6. Thus, p can be

1 or 2, and q can be 1, 2, 3, or 6.

The possible rational zeros, are

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

12

Example 3

USING THE RATIONAL ZEROS

THEOREM

Consider the polynomial function.

(b) Find all rational zeros and factor (x) into

linear factors.

Solution Use the remainder theorem to show

that 1 is a zero.

Use “trial and

error” to find

zeros.

(1) = 0

The 0 remainder shows that 1 is a zero. The quotient is

6×3 +13×2 + x – 2, so (x) = (x – 1)(6×3 +13×2 + x – 2).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

13

Example 3

USING THE RATIONAL ZEROS

THEOREM

Consider the polynomial function.

(b) Find all rational zeros and factor (x) into

linear equations.

Solution Now, use the quotient polynomial

and synthetic division to find that – 2 is a zero.

(– 2 ) = 0

The new quotient polynomial is 6×2 + x – 1.

Therefore, (x) can now be completely factored.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

14

Example 3

USING THE RATIONAL ZEROS

THEOREM

Consider the polynomial function.

(b) Find all rational zeros and factor (x) into

linear equations.

Solution

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

15

Example 3

USING THE RATIONAL ZEROS

THEOREM

Consider the polynomial function.

(b) Find all rational zeros and factor (x) into

linear equations.

Solution Setting 3x – 1 = 0 and 2x + 1 = 0

yields the zeros ⅓ and – ½. In summary the

rational zeros are 1, – 2, ⅓, and – ½. The linear

factorization of (x) is

Check by

multiplying

these factors.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

16

Note In Example 3, once we obtained

the quadratic factor 6×2 + x – 1, we were

able to complete the work by factoring it

directly. Had it not been easily factorable,

we could have used the quadratic formula

to find the other two zeros (and factors).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

17

Caution The rational zeros theorem gives only

possible rational zeros. It does not tell us whether

these rational numbers are actual zeros. We must rely

on other methods to determine whether or not they are

indeed zeros. Furthermore, the function must have

integer coefficients. To apply the rational zeros theorem

to a polynomial with fractional coefficients, multiply

through by the least common denominator of all the

fractions. For example, any rational zeros of p(x) defined

below will also be rational zeros of q(x).

Multiply the terms of

p(x) by 6.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

18

Fundamental Theorem of

Algebra

Every function defined by a polynomial

of degree 1 or more has at least one

complex zero.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

19

Fundamental Theorem of Algebra

From the fundamental theorem, if (x) is of

degree 1 or more, then there is some

number k1 such that f(k1) = 0. By the factor

theorem,

for some polynomial q1(x).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

20

Fundamental Theorem of Algebra

If q1(x) is of degree 1 or more, the

fundamental theorem and the factor

theorem can be used to factor q1(x) in the

same way. There is some number k2 such

that q1(k2) = 0, so

and

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

21

Fundamental Theorem of Algebra

Assuming that (x) has degree n and

repeating this process n times gives

where a is the leading coefficient of (x).

Each of these factors leads to a zero of (x),

so (x) has the same n zeros k1, k2, k3,…, kn.

This result suggests the number of zeros

theorem.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

22

Number of Zeros Theorem

A function defined by a polynomial of

degree n has at most n distinct zeros.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

23

Example 4

FINDING A POLYNOMIAL FUNCTION

THAT SATISFIES GIVEN CONDITIONS

(REAL ZEROS)

Find a polynomial function (x) of degree 3 with real

coefficients that satisfies the given conditions.

(a) Zeros of – 1, 2, and 4; (1) = 3

Solution These three zeros give

x – (– 1) = x + 1, x – 2, and x – 4 as factors of

(x). Since (x) is to be of degree 3, these

are the only possible factors by the number

of zeros theorem. Therefore, (x) has the

form

for some real number a.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

24

Example 4

FINDING A POLYNOMIAL FUNCTION

THAT SATISFIES GIVEN CONDITIONS

(REAL ZEROS)

Find a polynomial function (x) of degree 3 with real

coefficients that satisfies the given conditions.

(a) Zeros of – 1, 2, and 4; (1) = 3

Solution To find a, use the fact that (1) = 3.

Let x = 1.

(1) = 3

Multiply.

Divide by 6.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

25

Example 4

FINDING A POLYNOMIAL FUNCTION

THAT SATISFIES GIVEN CONDITIONS

(REAL ZEROS)

Find a polynomial function (x) of degree 3 with real

coefficients that satisfies the given conditions.

(a) Zeros of – 1, 2, and 4; (1) = 3

Solution Thus,

or

Multiply.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

26

Example 4

FINDING A POLYNOMIAL FUNCTION

THAT SATISFIES GIVEN CONDITIONS

(REAL ZEROS)

Find a polynomial function (x) of degree 3 with real

coefficients that satisfies the given conditions.

(b) – 2 is a zero of multiplicity 3; (– 1) = 4

Solution The polynomial function defined

by (x) has the following form.

Factor theorem

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

27

FINDING A POLYNOMIAL FUNCTION

THAT SATISFIES GIVEN CONDITIONS

(REAL ZEROS)

Example 4

Find a polynomial function (x) of degree 3 with real

coefficients that satisfies the given conditions.

(b) – 2 is a zero of multiplicity 3; (– 1) = 4

Solution To find a, use the fact that (–1) = 4.

Remember:

(x + 2)3 ≠ x3 + 23

Thus

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

28

Note In Example 4a, we cannot clear

the denominators in (x) by multiplying

each side by 2 because the result would

equal 2 • (x), not (x).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

29

Properties of Conjugates

For any complex numbers c and d, the

following properties hold.

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30

Conjugate Zeros Theorem

If (x) defines a polynomial function

having only real coefficients and if

z = a + bi is a zero of (x), where a

and b are real numbers, then

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

31

Proof of the Conjugate Zeros Theorem

Start with the polynomial function

where all coefficients are real numbers. If

the complex number z is a zero of (x), then

Take the conjugate of both sides of this

equation.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

32

Proof of the Conjugate Zeros Theorem

Using generalizations of the properties

Now use the property

that for any real number a,

and the fact

Hence is also a zero of (x),

which completes the proof.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

33

Caution When the conjugate zeros

theorem is applied, it is essential that

the polynomial have only real

coefficients. For example,

(x) = x – (1 + i)

has 1 + i as a zero, but the conjugate 1 – i

is not a zero.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

34

Example 5

FINDING A POLYNOMIAL FUNCTION THAT

SATISFIES GIVEN CONDITIONS (COMPLEX

ZEROS)

Find a polynomial function f(x) of least

degree having only real coefficients and

zeros 3 and 2 + i.

Solution The complex number 2 – i must

also be a zero, so the polynomial has at

least three zeros: 3, 2 + i, and 2 – i. For the

polynomial to be of least degree, these must

be the only zeros. By the factor theorem

there must be three factors, x – 3, x – (2 + i),

and x – (2 – i).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

35

Example 5

FINDING A POLYNOMIAL FUNCTION THAT

SATISFIES GIVEN CONDITIONS (COMPLEX

ZEROS)

Find a polynomial function f(x) of least

degree having only real coefficients and

zeros 3 and 2 + i.

Solution

Remember:

i2 = – 1

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

36

Example 5

FINDING A POLYNOMIAL FUNCTION THAT

SATISFIES GIVEN CONDITIONS (COMPLEX

ZEROS)

Find a polynomial function of least degree

having only real coefficients and zeros 3 and

2 + i.

Solution

Any nonzero multiple of

x3 – 7×2 + 17x – 15 also satisfies the given

conditions on zeros. The information on

zeros given in the problem is not sufficient to

give a specific value for the leading

coefficient.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

37

Example 6

FINDING ALL ZEROS GIVEN ONE

ZERO

Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,

given that 1 – i is a zero.

Solution Since the polynomial function has

only real coefficients and since 1 – i is a

zero, by the conjugate zeros theorem 1 + i is

also a zero. To find the remaining zeros,

first use synthetic division to divide the

original polynomial by x – (1 – i).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

38

Example 6

FINDING ALL ZEROS GIVEN ONE

ZERO

Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,

given that 1 – i is a zero.

Solution

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

39

Example 6

FINDING ALL ZEROS GIVEN ONE

ZERO

Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,

given that 1 – i is a zero.

Solution By the factor theorem, since

x = 1 – i is a zero of (x), x – (1 – i) is a

factor, and (x) can be written as follows.

We know that x = 1 + i is also a zero of (x).

Continue to use synthetic division and divide the

quotient polynomial above by x – (1 + i).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

40

Example 6

FINDING ALL ZEROS GIVEN ONE

ZERO

Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,

given that 1 – i is a zero.

Solution

Now f(x) can be written in the following factored

form.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

41

Example 6

FINDING ALL ZEROS GIVEN ONE

ZERO

Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,

given that 1 – i is a zero.

Solution

The remaining zeros are 2 and 3. The four

zeros are 1 – i, 1 + i, 2, and 3.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

42

Descartes’ Rule of Signs

Let (x) define a polynomial function with real coefficients

and a nonzero constant term, with terms in descending

powers of x.

(a) The number of positive real zeros of either equals the

number of variations in sign occurring in the coefficients of

(x), or is less than the number of variations by a positive

even integer.

(b) The number of negative real zeros of either equals

the number of variations in sign occurring in the coefficients

of (– x), or is less than the number of variations by a

positive even integer.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

43

APPLYING DESCARTES’ RULE OF

SIGNS

Example 7

Determine the different possibilities for the

number of positive, negative, and nonreal

complex zeros of

Solution We first consider the possible

number of positive zeros by observing that

(x) has three variations in signs:

1

2

3

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

44

Example 7

APPLYING DESCARTES’ RULE OF

SIGNS

Determine the different possibilities for the

number of positive, negative, and nonreal

complex zeros of

Solution Thus, by Descartes’ rule of signs,

has either 3 or 3 – 2 = 1 positive real zeros.

For negative zeros, consider the variations in

signs for (– x):

1

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

45

Example 7

APPLYING DESCARTES’ RULE OF

SIGNS

Determine the different possibilities for the

number of positive, negative, and nonreal

complex zeros of

Solution

1

Since there is only one variation in sign, (x)

has exactly one negative real zero.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

46

3

Polynomial

and Rational

Functions

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

1

3.2 Synthetic Division

• Synthetic Division

• Remainder Theorem

• Potential Zeros of Polynomial Functions

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

2

Division Algorithm

Let (x) and g(x) be polynomials with g(x) of

lesser degree than (x) and g(x) of degree 1 or

more. There exist unique polynomials q(x) and

r(x) such that

where either r(x) = 0 or the degree of r(x) is less

than the degree of g(x).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

3

Synthetic Division

Synthetic division provides an efficient

process for dividing a polynomial by a

binomial of the form x – k.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

4

Synthetic Division

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

5

Synthetic Division

Here the division

process is simplified

by omitting all

variables and writing

only coefficients, with

0 used to represent

the coefficient of any

missing terms.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

6

Synthetic Division

The numbers

in color that are

repetitions of

the numbers

directly above

them can be

omitted as

shown here.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

7

Synthetic Division

The entire process can now be condensed vertically.

The top row of numbers can be omitted since it

duplicates the bottom row if the 3 is brought down.

The rest of the bottom row is obtained by subtracting

– 12, – 40, and – 160 from the corresponding terms

above them.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

8

Synthetic Division

Additive

inverse

Signs

changed

Remainder

Quotient

To simplify the arithmetic, we replace subtraction in

the second row by addition and compensate by

changing the – 4 at the upper left to its additive

inverse, 4.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

9

Caution To avoid errors, use 0 as

the coefficient for any missing terms,

including a missing constant, when

setting up the division.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

10

Example 1

USING SYNTHETIC DIVISION

Use synthetic division to divide.

Solution Express x + 2 in the form x – k by

writing it as x – (– 2).

Coefficients

x + 2 leads

to – 2

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

11

Example 1

USING SYNTHETIC DIVISION

Use synthetic division to divide

Solution Bring down the 5, and multiply:

– 2(5) = – 10

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

12

Example 1

USING SYNTHETIC DIVISION

Use synthetic division to divide

Solution Add – 6 and – 10 to obtain – 16.

Multiply – 2(– 16) = 32.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

13

Example 1

USING SYNTHETIC DIVISION

Use synthetic division to divide

Solution Add – 28 and 32, obtaining 4.

Finally, – 2(4) = – 8.

Add columns.

Be careful

with signs.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

14

Example 1

USING SYNTHETIC DIVISION

Use synthetic division to divide

Solution Add – 2 and – 8 to obtain – 10.

Remainder

Quotient

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

15

Example 1

USING SYNTHETIC DIVISION

Since the divisor x – k has degree 1, the

degree of the quotient will always be written

one less than the degree of the polynomial to

be divided. Thus,

Remember to add

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

16

Special Case of the Division

Algorithm

For any polynomial (x) and any

complex number k, there exists a

unique polynomial q(x) and number r

such that the following holds.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

17

For Example

The mathematical statement

illustrates the special case of the division

algorithm. This form of the division algorithm is

useful in developing the remainder theorem.

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18

Remainder Theorem

If the polynomial (x) is divided by

x – k, the remainder is equal to (k).

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

19

Remainder Theorem

In Example 1, when (x) = 5×3 – 6×2 – 28x – 2

was divided by x + 2, or x – (– 2), the remainder

was – 10. Substitute – 2 for x in (x).

Use parentheses

around substituted

values to avoid errors.

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

20

Remainder Theorem

An alternative way to find the value of a

polynomial is to use synthetic division. By the

remainder theorem, instead of replacing x by – 2

to find (– 2), divide (x) by x + 2 using synthetic

division as in Example 1. Then (– 2) is the

remainder, – 10.

(– 2)

Copyright © 2017, 2013, 2009 Pearson Education, Inc.

21

Example 2

APPLYING THE REMAINDER

THEOREM

Let (x) = – x …

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