College Algebra
3.6
Real Zeros of a Polynomial Function
Objectives
• Use the Remainder and Factor Theorems
•Use the Rational Zeros Theorem
• Find the Real Zeros of a Polynomial Function
• Solve Polynomial Equations
•Use the Intermediate Value Theorem
Division Algorithm for Polynomials
If f(x) and g(x) denote polynomial functions and if
g(x) is not the zero polynomial, then there are unique
polynomial functions q(x) and r(x) such that
f (x)
r(x)
 q(x) 
g(x)
g(x)
dividend
or
quotient
f (x)  q(x)g(x)  r(x)
divisor
remainder
Remainder Theorem
Let f be a polynomial function. If f (x) is divided by
x - c, then the remainder is f (c).
Find the remainder if f ( x )  3x 3 9 x 2 - 18 x - 24
is divided by x + 3.
x + 3 = x - (-3)
f (-3)  3(-3)  9(-3) - 18(-3) - 24  30
3
2
Factor Theorem
Let f be a polynomial function. Then x - c is a
factor of f (x) if and only if f (c) = 0.
In other words, if f(c) = 0, then the remainder
found if f(x) is divided by x - c is zero.
Hence, since x - c divided into f(x) evenly
(remainder 0), x - c is a factor of f(x).
Use the Factor Theorem to determine whether the
3 9 x 2 - 18 x - 24
f
(
x
)
3
x

has the factor
function
(a) x + 3
(b) x + 4
a.) f(-3) = 30
Therefore, x + 3 does not divide into f.
So, x + 3 is not a factor of f.
b.) f(-4) = 0
Therefore, x + 4 does divide into f.
So, x + 4 is a factor of f.
Theorem: Number of Zeros
A polynomial function cannot have more
zeros than its degree.
Discuss the real zeros of
3
2
f ( x )  2 x  3x - 23x - 12
There are at most three zeros, since the function is
a polynomial of degree three.
Rational Zeros Theorem
Let f be a polynomial function of degree 1 or
higher of the form
f ( x )  a n x  a n -1 x
n
n -1
 a1 x  a0
where each coefficient is an integer. If p/q, in
lowest terms, is a rational zero of f, then p
must be a factor of a0 and q must be a factor
of an.
List the potential rational zeros of
f ( x)  2x3 3x2- 23x - 12
According the theorem, the numerator of potential
rational zeros will be factors of p = - 12 and the
denominator will be factors of q = 2
Factors of p:  1,  2,  3,  4,  6,  12
Factors of q:  1,  2
Potential
Rational Zeros
p
1 3
:  1,  2,  3,  4,  6,  12, , 
q
2 2
assignment

Page 379: 1-31 odds
5
4
3
2
Find the real zeros of f (x)  x  x - 9x - x  20x -12
Factor f over the real numbers.
First, determine the nature of the zeros.
Since the polynomial is degree 5, there are at most
five zeros.
Now, list all possible rational zeros p/q by factoring
the first and last coefficients of the function.
f (x)  x5 x4- 9x3- x2 20x -12
p:  1,  2,  3,  4,  6,  12
q: 1
p
:  1,  2,  3,  4,  6,  12
q
Now, graph the function and look for the
potential zeros.
f (x)  x5 x4- 9x3- x2 20x -12
Test k = -3
-3 1
1 - 9 -1
-3
6
20 -12
9 - 24
1 -2 -3
8
12
-4
0
Thus, -3 is a zero of f and x + 3 is a factor of f.
f ( x )  ( x  3)( x 4 - 2 x 3 - 3x 2  8 x - 4)
Test k = -2 - 2 1
1
-2 -3
8
-4
-2
8
-10
4
-4
5
-2
0
Thus, -2 is a zero of f and x + 2 is a factor of f.
f ( x )  ( x  3)( x  2 )( x 3 - 4 x 2  5 x - 2 )
f ( x )  ( x  3)( x  2 )( x - 4 x  5 x - 2 )
3
Test k = 1
1 1
1
2
-4
5
-2
1
-3
2
-3
2
0
Thus, 1 is a zero of f and x - 1 is a factor of f.
f ( x )  ( x  3)( x  2)( x - 1)( x - 3x  2)
2
f ( x )  ( x  3)( x  2 )( x - 1)( x - 1)( x - 2 )
f ( x )  ( x  3)( x  2 )( x - 1) ( x - 2 )
2
Theorem: Every polynomial function (with real
coefficients) can be uniquely factored into a product of
linear factors and/or irreducible quadratic factors.
In the previous example, while finding the real zeros of
the polynomial, f(x) was factored as follows:
5
4
3
2
f ( x )  x  x - 9 x - x  20 x - 12
f ( x )  ( x  3)( x  2 )( x - 1) 2 ( x - 2 )
The function was factored into four uniquely factors,
one of which had multiplicity 2.
Intermediate Value Theorem
Let f denote a continuous function. If a < b and if
f(a) and f(b) are of opposite sign, then the graph of f
has at least one x-intercept between a and b.
y
f(b)
x-intercept
a
f(a)

f (a )


f (b)

b
x
Use the intermediate value theorem to show that the
graph of the function
3
2
f ( x )  x  0 .2 x - 5 x - 1
has an x - intercept in the interval [-3, -2]
f (- 3) = -11.2 < 0
f (-2) = 1.8 > 0
Therefore, since f (-3) < 0 and f (-2) > 0, there
exists a number between - 3 and -2 where f (x) = 0.
So, the function has an x - intercept in the interval
[-3,-2]