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CHAPTER 4:
Polynomial and
Rational Functions
4.1 Polynomial Functions and Models
4.2 Graphing Polynomial Functions
4.3 Polynomial Division; The Remainder and Factor Theorems
4.4 Theorems about Zeros of Polynomial Functions
4.5 Rational Functions
4.6 Polynomial and Rational Inequalities
Copyright © 2009 Pearson Education, Inc.
4.4
Theorems about Zeros of Polynomial
Functions



Find a polynomial with specified zeros.
For a polynomial function with integer coefficients,
find the rational zeros and the other zeros, if possible.
Use Descartes’ rule of signs to find information
about the number of real zeros of a polynomial
function with real coefficients.
Copyright © 2009 Pearson Education, Inc.
The Fundamental Theorem of Algebra
Every polynomial function of degree n,
with n  1, has at least one zero in the
system of complex numbers.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-4
The Fundamental Theorem of Algebra
Example: Find a polynomial function of degree 4 having
zeros 1, 2, 4i, and 4i.
Solution: Such a polynomial has factors (x  1),(x  2),
(x  4i), and (x + 4i), so we have:
f ( x)  an ( x  1)( x  2)( x  4i)( x  4i)
Let an = 1: f ( x)  ( x  1)( x  2)( x  4i )( x  4i )
 ( x 2  3x  2)( x 2  16)
 x 4  3x3  2 x 2  16 x 2  48 x  32
 x 4  3x3  18 x 2  48 x  32
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-5
Zeros of Polynomial Functions with Real
Coefficients
Nonreal Zeros: If a complex number a + bi, b  0, is
a zero of a polynomial function f(x) with real
coefficients, then its conjugate, a  bi, is also a zero.
(Nonreal zeros occur in conjugate pairs.)
Irrational Zeros: If a  c b , where a, b, and c are
rational and b is not a perfect square, is a zero of a
polynomial function f(x) with rational coefficients,
then its conjugate a  c b , is also a zero.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-6
Example
Suppose that a polynomial function of degree 6 with
rational coefficients has 3 + 2i, 6i, and 1  2 as
three of its zeros. Find the other zeros.
Solution: The other zeros are the conjugates of the
given zeros, 3  2i, 6i, and 1  2. There are no
other zeros because the polynomial of degree 6 can
have at most 6 zeros.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-7
Rational Zeros Theorem
Let
P( x)  an x n  an 1 x n 1  ...  a2 x 2  a1 x  a0 ,
where all the coefficients are integers. Consider a
rational number denoted by p/q, where p and q are
relatively prime (having no common factor besides
1 and 1). If p/q is a zero of P(x), then p is a factor
of a0 and q is a factor of an.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-8
Example
Given f(x) = 2x3  3x2  11x + 6:
a) Find the rational zeros and then the other zeros.
b) Factor f(x) into linear factors.
Solution:
a) Because the degree of f(x) is 3, there are at most 3
distinct zeros. The possibilities for p/q are:
Possibilities for p
:
Possibilities for q
1, 2, 3, 6
1, 2
Possibilities for p / q :
1, 1, 2, 2,3, 3,6, 6, 12 , 21 , 23 , 23
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-9
Example continued
Use synthetic division to help determine the zeros. It is easier
to consider the integers before the fractions.
We try 1:
We try 1:
1 2 –3 –11
6
2 –1 –12
2 –1 –12 –6
Since f(1) = 6, 1 is
not a zero.
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–1 2 –3 –11
–2
2 –5
6
5 6
–6 12
Since f(1) = 12, 1 is
not a zero.
Slide 4.4-10
Example continued
We try 3:
3 2 –3 –11 6
6
9 –6
2 3 –2 0
Since f(3) = 0, 3 is a zero. Thus x  3 is a factor. Using the
results of the division above, we can express f(x) as
f ( x)  ( x  3)(2 x 2  3x  .2)
We can further factor 2x2 + 3x  2 as (2x  1)(x + 2).
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-11
Example continued
The rational zeros are 2, 3 and 1 .
2
The complete factorization of f(x) is:
f ( x)  (2 x  1)( x  3)( x  2)
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-12
Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and a
nonzero constant term. The number of positive real zeros of
P(x) is either:
1. The same as the number of variations of sign in P(x), or
2. Less than the number of variations of sign in P(x) by a
positive even integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(x), or
4. Less than the number of variations of sign in P(x) by a
positive even integer.
A zero of multiplicity m must be counted m times.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-13
Example
What does Descartes’ rule of signs tell us about the
number of positive real zeros and the number of
negative real zeros?
P( x)  3x 4  6 x 3  x 2  7 x  2
 3x 4  6 x3  x 2  7 x  2
There are two variations of sign, so there are either
two or zero positive real zeros to the equation.
Copyright © 2009 Pearson Education, Inc.
Slide 4.4-14
Example continued
P( x)  3( x) 4  6( x)3  ( x) 2  7( x)  2
 3x 4  6 x3  x 2  7 x  2
There are two variations of sign, so there are either
two or zero negative real zeros to the equation.
Total Number of Zeros = 4:
Positive
2
2
0
Negative
2
0
2
Nonreal
0
2
2
Copyright © 2009 Pearson Education, Inc.
0
0
4
Slide 4.4-15