* Evaluate a polynomial using the Remainder Theorem. * Use the Factor Theorem to solve a polynomial equation. * Use the Rational Zero Theorem to find rational zeros. * Find zeros of a polynomial function. * Use the Linear Factorization Theorem to find polynomials with given zeros. * Use Descartes’ Rule of Signs.

1. 5.5 Zeros of Polynomial Functions
Chapter 5 Polynomial and Rational Functions

2. Concepts and Objectives
⚫ Objectives for this section are
⚫ Evaluate a polynomial using the Remainder Theorem.
⚫ Use the Factor Theorem to solve a polynomial
equation.
⚫ Use the Rational Zero Theorem to find rational zeros.
⚫ Find zeros of a polynomial function.
⚫ Use the Linear Factorization Theorem to find
polynomials with given zeros.
⚫ Use Descartes’ Rule of Signs.

3. Remainder Theorem
⚫ The remainder theorem:
⚫ Example: Let . Find f(–3).
If the polynomial f(x) is divided by x – k, then the
remainder is equal to f(k).
( )= − + − −
4 2
3 4 5
f x x x x

4. Remainder Theorem
⚫ The remainder theorem:
⚫ Example: Let . Find f(–3).
If the polynomial f(x) is divided by x – k, then the
remainder is equal to f(k).
( )= − + − −
4 2
3 4 5
f x x x x
− − − −
3 1 0 3 4 5
–1
3
3
–9
–6
18
14
–42
–47
( )
− = −
3 47
f

5. Factor Theorem
⚫ Example: Determine whether x + 4 is a factor of
The polynomial x – k is a factor of the polynomial
f(x) if and only if f(k) = 0.
( )= − + +
4 2
3 48 8 32
f x x x x

6. Factor Theorem
⚫ Example: Determine whether x + 4 is a factor of
Yes, it is.
The polynomial x – k is a factor of the polynomial
f(x) if and only if f(k) = 0.
( )= − + +
4 2
3 48 8 32
f x x x x
− −
4 3 0 48 8 32
–12
3
0
0
48
–12
0
–32
8

7. Rational Zeros Theorem
In other words, the numerator is a factor of the constant
term and the denominator is a factor of the first
coefficient.
If p/q is a rational number written in lowest
terms, and if p/q is a zero of f, a polynomial
function with integer coefficients, then p is a
factor of the constant term and q is a factor of
the leading coefficient.

8. Rational Zeros Theorem (cont.)
⚫ Example: For the polynomial function defined by
(a) List all possible rational zeros
(b) Find all rational zeros and factor f(x) into linear
factors.
( )= − − + +
4 3 2
8 26 27 11 4
f x x x x x

9. Rational Zeros Theorem (cont.)
⚫ Example: For the polynomial function defined by
(a) List all possible rational zeros
For a rational number to be zero, p must be a
factor of 4 and q must be a factor of 8:
( )= − − + +
4 3 2
4
26 27 1
8 1
f x x x x x
 
   
1, 2, 4
p
p
q
 
      
 
 
1 1 1
1, 2, 4, , ,
2 4 8
p
q
 
    
, 1, 2, 4, 8
q

10. Rational Zeros Theorem (cont.)
⚫ Example: For the polynomial function defined by
(b) Find all rational zeros and factor f(x) into linear
factors.
Look at the graph of f(x) to judge where it crosses
the x-axis:
( )= − − + +
4 3 2
8 26 27 11 4
f x x x x x

11. Rational Zeros Theorem (cont.)
⚫ Example: For the polynomial function defined by
Use synthetic division to show that –1 is a zero:
(If you don’t have a graph of the function, you will need
to use guess-and-check on the possible factors.)
( )= − − + +
4 3 2
8 26 27 11 4
f x x x x x
− − −
1 8 26 27 11 4
–8
8 –34
34
7
–7
4
–4
0
( ) ( )( )
= + − + +
3 2
1 8 34 7 4
f x x x x x

12. Rational Zeros Theorem (cont.)
⚫ Example, cont.
Now, we can check the remainder for a zero at 4:
zeros are at –1, 4,
( ) ( )( )
= + − + +
3 2
1 8 34 7 4
f x x x x x
−
4 8 34 7 4
32
8 –2
–8
–1
–4
0
( ) ( )( )( )
= + − − −
2
1 4 8 2 1
f x x x x x
( ) ( )( )( )( )
= + − + −
1 4 4 1 2 1
f x x x x x
−
1 1
,
4 2

13. Fundamental Theorem of Algebra
The number of times a zero occurs is referred to as the
multiplicity of the zero.
Every function defined by a polynomial of degree
1 or more has at least one complex zero.
A function defined by a polynomial of degree n
has at most n distinct zeros.

14. Fundamental Theorem of Algebra
⚫ Example: Find a function f defined by a polynomial of
degree 3 that satisfies the following conditions.
(a) Zeros of –3, –2, and 5; f(–1) = 6
(b) 4 is a zero of multiplicity 3; f(2) = –24

15. Fundamental Theorem of Algebra
⚫ Example: Find a function f defined by a polynomial of
degree 3 that satisfies the following conditions.
(a) Zeros of –3, –2, and 5; f(–1) = 6
Since f is of degree 3, there are at most 3 zeros, so these
three must be it. Therefore, f(x) has the form
( ) ( )( )( )
= + + −
3 2 5
f x a x x x
( ) ( )
( ) ( )
( )( )
= − − − − −
3 2 5
f x a x x x

16. Fundamental Theorem of Algebra
⚫ Example, cont.
We also know that f(–1) = 6, so we can solve for a:
Therefore, or
( ) ( )( )( )
− −
= −
− −
+ +
3
1 1 1 2 5
1
f a
( )( )( )
= − = −
6 2 1 6 12
a a
= −
1
2
a
( ) ( )( )( )
= − + + −
1
3 2 5
2
f x x x x
( )= − + +
3
1 19
15
2 2
f x x x

17. Fundamental Theorem of Algebra
⚫ Example: Find a function f defined by a polynomial of
degree 3 that satisfies the following conditions.
(b) 4 is a zero of multiplicity 3; f(2) = –24
This means that the zero 4 occurs 3 times:
or
( ) ( )( )( )
= − − −
4 4 4
f x a x x x
( ) ( )
= −
3
4
f x a x

18. Fundamental Theorem of Algebra
⚫ Example, cont.
Since f(2) = –24, we can solve for a:
Therefore, or
( ) ( )
= −
3
4
2 2
f a
( )
− = − = −
3
24 2 8
a a
=3
a
( ) ( )
= −
3
3 4
f x x
( )= − + −
3 2
3 36 144 192
f x x x x

19. Linear Factorization Theorem
⚫ A vital implication of the Fundamental Theorem of
Algebra is that a polynomial function of degree n will
have n zeros in the set of complex numbers, if we allow
for multiplicities. (Real numbers are also complex
numbers.)
If a polynomial function has degree n, then it will
have n factors, and each factor will be in the form
(x ‒ c), where c is a complex number.

20. Conjugate Conjugate Theorem
This means that if 3 + 2i is a zero for a polynomial
function with real coefficients, then it also has 3 – 2i as a
zero.
If f(x) defines a polynomial function having only
real coefficients and if z = a + bi is a zero of f(x),
where a and b are real numbers, then z = a – bi
is also a zero of f(x).

21. Conjugate Conjugate Theorem
⚫ Example: Find a polynomial function of least degree
having only real coefficients and zeros –4 and 3 – i.

22. Conjugate Conjugate Theorem
⚫ Example: Find a polynomial function of least degree
having only real coefficients and zeros –4 and 3 – i.
The complex number 3 + i must also be a zero, so the
polynomial has at least three zeros and has to be at least
degree 3. We don’t know anything else about the
function, so we will let a = 1.
( ) ( )
( ) ( )
( ) ( )
( )
= − − − − − +
4 3 3
f x x x i x i
( ) ( ) ( ) ( ) ( )( )
( )
= + − + − − + − +
2
4 3 3 3 3
f x x x i x i x i i
( ) ( )( )
2 2
4 3 3 9
f x x x x ix x ix i
= + − − − + + −
( ) ( )( )
= + − + = − − +
2 3 2
4 6 10 2 14 40
f x x x x x x x

23. Descartes’ Rule of Signs
If f(x) defines a polynomial function having
only real coefficients:
• The number of positive real zeros is either
equal to the number of sign changes of f(x)
or is less than the number of sign changes by
an even integer.
• The number of negative real zeros is either
equal to the number of sign changes of f(‒x)
or is less than the number of sign changes by
an even integer.

24. Descartes’ Rule of Signs
⚫ Example:
The function must have 1 positive real zero
f(x) could have 3 or 1 negative real zeros (3‒2=1)
( ) 4 3 2
1
f x x x x x
= + + + −
sign change
( ) ( ) ( ) ( ) ( )
4 3 2
4 3 2
1
1
f x x x x x
x x x x
− = − + − + − + − −
= + − + − −

25. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
3 2
13 57 85 0
x x x
+ + + =

26. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
First, we use the Rational Zero Theorem to find possible
real solutions:
factors of 85: ±1, ±5, ±17, ±85; factors of 1: ±1
3 2
13 57 85 0
x x x
+ + + =
factor of constant term factor of 85
factor of leading coeff. factor of 1
p
q
= =
1, 5, 17, 85
p
q
=    

27. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
Now, we will test the possible solutions until we find
one with a remainder of 0.
So 1 is not a solution.
3 2
13 57 85 0
x x x
+ + + =
1 1 13 57 85
1 14 43
1 14 43 128

28. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
Now, we will test the possible solutions until we find
one with a remainder of 0.
So ‒5 is a solution, and the quotient is .
3 2
13 57 85 0
x x x
+ + + =
‒5 1 13 57 85
‒5 ‒40 ‒85
1 8 17 0
2
8 17
x x
+ +

29. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
Now, we can use the Quadratic Formula to find the
other two solutions:
3 2
13 57 85 0
x x x
+ + + =
( )( )
( )
2
8 8 4 1 17
2 1
8 4 8 2
2 2
4
x
i
i
−  −
=
−  − − 
= =
= − 

30. Putting It All Together
⚫ Example: Find all complex solutions (real and non-real).
So, the zeros of this equation are ‒5, ‒4+i, and ‒4‒i.
3 2
13 57 85 0
x x x
+ + + =

31. Classwork
⚫ College Algebra 2e
⚫ 5.5: 6-20 (even); 5.4: 38-52 (even); 5.3: 48-76 (4)
⚫ 5.5 Classwork Check
⚫ Quiz 5.4