This document discusses key concepts related to finding zeros of polynomial functions including: 1) The factor theorem, which states that a polynomial x - k is a factor of a function f(x) if and only if f(k) = 0. 2) The rational zeros theorem, which gives possible rational zeros based on the factors of the leading coefficient and constant term. 3) The fundamental theorem of algebra, which states that every polynomial of degree n has n complex zeros and examples of finding functions based on given zeros. 4) The conjugate zeros theorem, which states that if z = a + bi is a zero, then z = a - bi is also a zero for polynomials with real coefficients. 5

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

2. Concepts and Objectives
 Zeros of Polynomial Functions
 Find rational zeros of a polynomial function
 Use the Fundamental Theorem of Algebra to find a
function that satisfies given conditions
 Find all zeros of a polynomial function

3. 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 32f x x x x

4. 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 32f x x x x
 4 3 0 48 8 32
–123
0
0
48–12
0
–32
8

5. 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.

6. 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 4f x x x x x

7. 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
8 26 27 11 4f x x x x x
    1, 2, 4p
p
q
 
       
 
1 1 1
1, 2, 4, , ,
2 4 8
p
q
     , 1, 2, 4, 8q

8. 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 4f x x x x x

9. Rational Zeros Theorem (cont.)
 Example: For the polynomial function defined by
Use synthetic division to show that –1 is a zero:
     4 3 2
8 26 27 11 4f 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 4f x x x x x

10. 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 4f x x x x x
4 8 34 7 4
32
8 –2
–8
–1
–4
0
         2
1 4 8 2 1f x x x x x
          1 4 4 1 2 1f x x x x x

1 1
,
4 2

11. 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.

12. 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

13. 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 5f x a x x x
             3 2 5f x a x x x

14. Fundamental Theorem of Algebra
 Example, cont.
We also know that f–1 = 6, so we can solve for a:
Therefore, or
         31 1 1 2 51f a
      6 2 1 6 12a a
 
1
2
a
         
1
3 2 5
2
f x x x x
    31 19
15
2 2
f x x x

15. 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 4f x a x x x
    
3
4f x a x

16. Fundamental Theorem of Algebra
 Example, cont.
Since f2 = –24, we can solve for a:
Therefore, or
    
3
42 2f a
     
3
24 2 8a a
3a
    
3
3 4f x x
    3 2
3 36 144 192f x x x x

17. Conjugate Zeros 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.

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

19. Conjugate Zeros 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 3f x x x i x i
                   2
4 3 3 3 3f x x x i x i x i i
    2 2
4 3 3 9f x x x x ix x ix i       
           2 3 2
4 6 10 2 14 40f x x x x x x x

20. Putting It All Together
 Example: Find all zeros of
given that 2 + i is a zero.
     4 3 2
17 55 50f x x x x x

21. Putting It All Together
 Example: Find all zeros of
given that 2 + i is a zero.
First, we divide the function by :
     4 3 2
17 55 50f x x x x x
   2x i
   2 1 1 17 55 50i
1
2+i
1+i
1+3i
–16+3i
–35–10i
20–10i
50
0
                    3 2
2 1 16 3 20 10f x x i x i x i x i

22. Putting It All Together
 Example, cont.
We also know that the conjugate, 2 – i is a zero, so we
can divide the remainder by this:
                    3 2
2 1 16 3 20 10f x x i x i x i x i
    2 1 1 16 3 20 10i i i i
1
2–i
3
6–3i
–10
–20+10i
0
              2
2 2 3 10f x x i x i x x

23. Putting It All Together
 Example, cont.
Lastly, we can factor or use the quadratic formula to find
our remaining zeros:
So, our zeros are at 2+i, 2–i, –5, and 2.
              2
2 2 3 10f x x i x i x x
      2
3 10 5 2x x x x
               2 2 5 2f x x i x i x x

24. Classwork
 College Algebra
 Page 337: 6-22 (even), page 326: 28-40 (4),
page 318: 68-72 (even)