slide 3.4- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Zeros of a Polynomial Function

Learn to find the possible number of positive and negative zeros of polynomials.

Learn to find the bounds on the real zeros of polynomials.

Learn basic facts about the complex zeros of polynomials.

Learn to use the Conjugate Pairs Theorem to find zeros of polynomials.

SECTION 3.4

1

2

3

4


DESCARTE’S RULE OF SIGNS

Let F(x) be a polynomial function with real coefficients and with terms written in descending order.

1. The number of positive zeros of F is either equal to the number of variations of sign of F(x) or less than that number by an even integer.


DESCARTE’S RULE OF SIGNS

When using Descarte’s Rule, a zero of multiplicity m should be counted as m zeros.

2. The number of negative zeros of F is either equal to the number of variations of sign of F(–x) or less than that number by an even integer.


EXAMPLE 1 Using Descarte’s Rule of Signs

Find the possible number of positive and negative zeros of

f x x5 3x3 5x2 9x 7.

Solution

There are three variations in sign in f (x).

f x x5 3x3 5x2 9x 7

+ to – – to + + to –The number of positive zeros is either 3 or (3 – 2 =) 1.


EXAMPLE 1 Using Descarte’s Rule of Signs

Solution continued

There are two variations in sign in f (–x).

f x x 5 3 x 3 5 x 2 9 x 7

x5 3x3 5x2 9x 7

+ to –– to +

The number of negative zeros is either 2 or (2 – 2 =) 0.


RULES FOR BOUNDS

Let F(x) be a polynomial function with real coefficients and a positive leading coefficient. Suppose F(x) is synthetically divided by x – k. Then

1. If k > 0, and each number in the last row is either zero or positive, then k is an upper bound on the zeros of F(x).


RULES FOR BOUNDS

If 0 appears in the last row of the synthetic division, then it may be assigned either a positive or a negative sign in determining whether the signs alternate.

2. If k < 0, and numbers in the last row alternate in sign, then k is a lower bound on the zeros of F(x).


EXAMPLE 2 Finding the Bounds on the Zeros

Find upper and lower bounds on the zeros of

F x x4 x3 5x2 x 6.SolutionThe possible zeros are: ±1, ±2, ±3, and ±6. There is only one variation in sign so there is one positive zero. Use synthetic division until last row is all positive or 0. Not 1, not 2, but 3. 3 1 1 5 1 6

3 6 3 6

1 2 1 2 0


EXAMPLE 2 Finding the Bounds on the Zeros

Solution continued

So 3 is upper bound.

Use synthetic division until last row alternates in sign. Not –1, but –2.

So –2 is lower bound.

2 1 1 5 1 6

2 6 2 6

1 3 1 3 0


FUNDAMENTAL THEOREM OF ALGEBRA

Every polynomial

with complex coefficients an, an – 1, …, a1, a0 has at least one complex zero.

P x an xn an 1xn 1 ... a1x a0 n 1, an 0


FACTORIZATION THEOREM FOR POLYNOMIALS

If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form

P x a x r1 x r2 ... x rn ,where a, r1, r2, … , rn are complex numbers.


NUMBER OF ZEROS THEOREM

A polynomial of degree n has exactly n zeros, provided that a zero of multiplicity k is counted k times.


EXAMPLE 3Constructing a Polynomial Whose Zeros are Given

Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and –i. Write P(x)

Solution

a. Since P(x) has degree 4, we writeP x a x r1 x r2 x r3 x r4

2 x 1 x 3 x i x i 2 x 1 x 3 x i x i

a. in completely factored form;b. by expanding the product found in part a.


EXAMPLE 3Constructing a Polynomial Whose Zeros are Given

Solution continued

b. by expanding the product found in part a.

P x 2 x 1 x 3 x i x i 2 x 1 x 3 x2 1 2 x 1 x3 3x2 x 3 2 x4 2x3 2x2 2x 3 2x4 4x3 4x2 4x 6


CONJUGATE PAIRS THEOREM

If P(x) is a polynomial function whose coefficients are real numbers and ifz = a + bi is a zero of P, then its conjugate,

is also a zero of P.z a bi,


ODD–DEGREE POLYNOMIALSWITH REAL ZEROS

Any polynomial P(x) of odd degree with real coefficients must have at least one real zero.


EXAMPLE 4 Using the Conjugate Pairs Theorem

A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x).

2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i

Solution

Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are:


FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS

Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or prime quadratic factors.


EXAMPLE 5Finding the Complex Real Zeros of a Polynomial

Given that 2 – i is a zero of

SolutionThe conjugate of 2 – i, 2 + i is also a zero.

P x x4 6x3 14x2 14x 5,

x 2 i x 2 i x 2 i x 2 i

x 2 2 i2 x2 4x 5

So P(x) has linear factors:

find the remaining zeros.

x 2 i x 2 i



Solution continued Divide P(x) by x2 – 4x + 5

x2 4x 5 x4 6x3 14x2 14x 5

x4 4x3 5x2

2x3 9x2 14x

2x3 8x2 10x

x2 4x 5

x2 4x 5

0

x2 2x 1



Solution continued

We can multiply these factors to find P(x).

P x x2 2x 1 x2 4x 5 x 1 x 1 x2 4x 5 x 1 x 1 x 2 i x 2 i

The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.


EXAMPLE 6 Finding the Zeros of a Polynomial

Find all zeros of the polynomial P(x) = x4 – x3 + 7x2 – 9x – 18.

Solution

Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18Use synthetic division to find that 2 is a zero.

2 1 1 7 9 18

2 2 18 18

1 1 9 9 0

(x – 2) is a factor of P(x). Solve x3 x2 9x 9 0


EXAMPLE 6 Finding the Zeros of a Polynomial

Solution continued

x3 x2 9x 9 0

x2 x 1 9 x 1 0

x 1 x2 9 0

x 1 0 or x2 9 0

x 1 or x2 9

x 1 or x 3i

The four zeros of P(x) are –1, 2, –3i, and 3i.

slide 3.4- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

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