3.3 Zeros of Polynomial Functions

Zeros of Polynomial Functions
Chapter 3 Polynomial and Rational Functions

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

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

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

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.

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

(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

(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

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

 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

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.

 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

(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

 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

(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

 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

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.

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

 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

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

 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

 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

 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

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

3.3 Zeros of Polynomial Functions

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3.3 Zeros of Polynomial Functions