Chapter 3. Polynomial and Rational Functions. 3.4 Zeros of Polynomial Functions

Содержание

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Use the Rational Zero Theorem to find possible rational zeros. Find

Use the Rational Zero Theorem to find possible rational zeros.
Find zeros

of a polynomial function.
Solve polynomial equations.
Use the Linear Factorization Theorem to find polynomials with given zeros.
Use Descartes’ Rule of Signs.

Objectives:

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The Rational Zero Theorem If has integer coefficients and (where is

The Rational Zero Theorem

If has
integer coefficients and (where is reduced

to
lowest terms) is a rational zero of f, then p is a factor of the constant term, a0, and q is a factor of the leading coefficient, an.
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Example: Using the Rational Zero Theorem List all possible rational zeros

Example: Using the Rational Zero Theorem

List all possible rational zeros of
The

constant term is –3 and the leading coefficient is 4.
Factors of the constant term, –3:
Factors of the leading coefficient, 4:
Possible rational zeros are:
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Example: Finding Zeros of a Polynomial Function Find all zeros of

Example: Finding Zeros of a Polynomial Function

Find all zeros of
We

begin by listing all possible rational zeros.
Possible rational zeros =
We now use synthetic division to see if we can find a rational zero among the four possible rational zeros.
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Example: Finding Zeros of a Polynomial Function (continued) Find all zeros

Example: Finding Zeros of a Polynomial Function (continued)

Find all zeros of


Possible rational zeros are 1, –1, 2, and –2. We will use synthetic division to test the possible rational zeros.
Neither –2 nor –1 is a zero. We continue testing possible rational zeros.
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Example: Finding Zeros of a Polynomial Function (continued) Find all zeros

Example: Finding Zeros of a Polynomial Function (continued)

Find all zeros of


Possible rational zeros are 1, –1, 2, and –2. We will use synthetic division to test the possible rational zeros. We have found that –2 and –1 are not rational zeros. We continue testing with 1 and 2.
We have found a rational zero at x = 2.
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Example: Finding Zeros of a Polynomial Function (continued) Find all zeros

Example: Finding Zeros of a Polynomial Function (continued)

Find all zeros of


We have found a rational zero at x = 2.
The result of synthetic division is:
This means that
We now solve
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Example: Finding Zeros of a Polynomial Function (continued) Find all zeros

Example: Finding Zeros of a Polynomial Function (continued)

Find all zeros of


We have found that
We now solve

The solution set is

The zeros of
are

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Properties of Roots of Polynomial Equations 1. If a polynomial equation

Properties of Roots of Polynomial Equations

1. If a polynomial equation is

of degree n, then counting multiple roots separately, the equation has n roots.
2. If a + bi is a root of a polynomial equation with real coefficients then the imaginary number
a – bi is also a root. Imaginary roots, if they exist, occur in conjugate pairs.
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Example: Solving a Polynomial Equation Solve We begin by listing all

Example: Solving a Polynomial Equation

Solve
We begin by listing all possible rational

roots:
Possible rational roots =
Possible rational roots are 1, –1, 13, and –13. We will use synthetic division to test the possible rational zeros.
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Example: Solving a Polynomial Equation (continued) Solve Possible rational roots are

Example: Solving a Polynomial Equation (continued)

Solve
Possible rational roots are 1, –1,

13, and –13. We will use synthetic division to test the possible rational zeros.
x = 1 is a root for this polynomial.
We can rewrite the equation in factored form
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Example: Solving a Polynomial Equation (continued) Solve We have found that

Example: Solving a Polynomial Equation (continued)

Solve
We have found that x =

1 is a root for this polynomial.
In factored form, the polynomial is
We now solve
We begin by listing all possible rational roots.
Possible rational roots =
Possible rational roots are 1, –1, 13, and –13. We will use synthetic division to test the possible rational zeros.
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Example: Solving a Polynomial Equation (continued) Solve Possible rational roots are

Example: Solving a Polynomial Equation (continued)

Solve
Possible rational roots are 1, –1,

13, and –13. We will use synthetic division to test the possible rational zeros. Because –1 did not work for the original polynomial, it is not necessary to test that value.
The factored form of this polynomial is

x = 1 is a (repeated) root
for this polynomial

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Example: Solving a Polynomial Equation (continued) Solve The factored form of

Example: Solving a Polynomial Equation (continued)

Solve
The factored form of this polynomial

is
We will use the quadratic formula to solve

The solution set of the original
equation is

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The Fundamental Theorem of Algebra If f(x) is a polynomial of

The Fundamental Theorem of Algebra

If f(x) is a polynomial of degree

n, where then the equation f(x) = 0 has at least one complex root.
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The Linear Factorization Theorem If where and then where c1, c2,

The Linear Factorization Theorem

If where
and then
where c1, c2, ..., cn

are complex numbers (possibly real and not necessarily distinct). In words: An nth-degree polynomial can be expressed as the product of a nonzero constant and n linear factors, where each linear factor has a leading coefficient of 1.
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Example: Finding a Polynomial Function with Given Zeros Find a third-degree

Example: Finding a Polynomial Function with Given Zeros

Find a third-degree polynomial

function f(x) with real coefficients that has –3 and i as zeros and such that
f(1) = 8.
Because i is a zero and the polynomial has real coefficients, the conjugate, –i, must also be a zero. We can now use the Linear Factorization Theorem.
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Example: Finding a Polynomial Function with Given Zeros Find a third-degree

Example: Finding a Polynomial Function with Given Zeros

Find a third-degree polynomial

function f(x) with real coefficients that has –3 and i as zeros and such that
f(1) = 8.
Applying the Linear Factorization Theorem, we found
that

The polynomial function is

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Descartes’ Rule of Signs Let be a polynomial with real coefficients.

Descartes’ Rule of Signs

Let
be a polynomial with real coefficients.
1. The number

of positive real zeros of f is either
a. the same as the number of sign changes of f(x)
or
b. less than the number of sign changes of f(x) by a
positive even integer. If f(x) has only one
variation in sign, then f has exactly one positive
real zero.
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Descartes’ Rule of Signs (continued) Let be a polynomial with real

Descartes’ Rule of Signs (continued)

Let
be a polynomial with real coefficients.
2. The

number of negative real zeros of f is either
a. The same as the number of sign changes in f(–x)
or
b. less than the number of sign changes in f(–x) by
a positive even integer. If f(–x) has only one
variation in sign, then f has exactly one negative
real zero
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Example: Using Descartes’ Rule of Signs Determine the possible number of

Example: Using Descartes’ Rule of Signs

Determine the possible number of positive

and negative real zeros of
1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x).
There are 4 variations in sign.
The number of positive real zeros of f is either 4, 2, or 0.