Transcript Algebra Expressions and Real Numbers

Section 2.5
Zeros
of
Polynomial Functions
Objectives:
• Use the Rational Zero Theorem to find
possible rational zeros
• Find zeros of a polynomial function
• Solve polynomial equations
• Use the Linear Factorization Theorem to
find polynomials with given zeros
• Use Descartes’s Rule of Signs
The Rational Zero Theorem
The Rational Zero Theorem - theorem that gives a list of possible
rational zeros of a polynomial function. Not every number in the list
will be a zero of the function, but every rational zero of the
polynomial function will appear somewhere in the list.
The Rational Zero Theorem
If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients
p
p
and q (where q is reduced to lowest terms) is a rational zero
of f, then p is a factor of the constant term, a0,
and q is a factor of the leading coefficient an.
Example 1
List all possible rational zeros of f(x) = 15x3 + 14x2 – 3x – 2.
Solution
The constant term is – 2 and the leading coefficient is 15.
Factors of the constant term,  2
Factors of the leading coefficient, 15
1,  2

1,  3,  5,  15
Possible rational zeros 
 1,  2,
 13 ,  23 ,
Divide
1
and 2
by 1.
Divide
1
and 2
by 3.
 15 ,  52 ,
Divide
1
and 2
by 5.
1,  2
 15
15
Divide
1
and 2
by 15.
There are 16 possible rational zeros.
The actual solution set to f(x) = 15x3 + 14x2 – 3x – 2 = 0 is {-1, -1/3, 2/5},
which contains 3 of the 16 possible solutions.
Example 2
1) List all possible rational zeros of
f ( x)  x3  3x2  4 x 12
2) Find one of the zeros of the function using synthetic division,
then factor the remaining polynomial.
3) What are all of the zeros of the function? How can the graph
below help you find the zeros?
Example 3
1)
List all possible rational zeros of
f ( x)  6 x3 19 x 2  2 x  3
2) Starting with the integers, find one zero of the function using
synthetic division, then factor the remaining polynomial.
3) What are all of the zeros of the function?
Example 4
1) List all possible rational roots of
x4  x3  7 x2  9 x 18  0
2) Starting with the integers, find two roots of the equation using
synthetic division. The graph below will help you easily find
those roots.
3) Factor the remaining polynomial. What are all of the roots of the
equation?
4) The graph below will NOT help you find the imaginary roots.
Why?
Example 5
Solve:
x4 – 6x2 – 8x + 24 = 0
1. Begin by listing all possible rational roots (zeros):
Factors of the constant term, 24
Factors of the leading coefficient, 1
1,  2  3,  4,  6,  8,  12,  24

  1,  2  3,  4,  6,  8,  12,  24
1
Possible rational zeros 
2. The graph of f (x) = x4 – 6x2 – 8x + 24 is shown the figure below.
Because the x-intercept is 2, we will test 2 by synthetic division
and show that it is a root of the given equation.
2
1
1
x-intercept: 2
0 6 8 24
2
4 4 24
2 2 12 0
The zero remainder
indicates that 2 is a root
of x4 - 6x2 - 8x + 24 = 0.
Example 5
Solve:
x4 – 6x2 – 8x + 24 = 0
3. Now we can rewrite the given equation in factored form.
x4 – 6x2 + 8x + 24 = 0
(x – 2)(x3 + 2x2 – 2x – 12) = 0
x–2=0
or
x3 + 2x2 – 2x – 12 = 0
This is the given equation.
This is the result obtained from the
synthetic division.
Set each factor equal to zero.
4. However, take a second look at the figure
of the graph of x4 – 6x2 – 8x + 24 = 0.
Because the graph turns around at 2,
this means that 2 is a root of even multiplicity.
So, 2 must also be a root of x3 + 2x2 – 2x – 12 = 0,
confirmed by the following synthetic division.
2
1
1
2 2 12
2 8 12
4 6
0
These are the coefficients of
x3 + 2x2 - 2x - 12 = 0.
The zero remainder indicates that 2 is
a root of x3 + 2x2 - 2x - 12 = 0.
Example 5
Solve:
x4 – 6x2 – 8x + 24 = 0
5. Now we can solve the original equation as follows.
x4 – 6x2– 8x + 24 = 0
(x – 2)(x3 + 2x2 – 2x – 12) = 0
(x – 2)(x – 2)(x2 + 4x + 6) = 0
x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0
x=2
x=2
x2 + 4x + 6 = 0
This is the given equation.
This was obtained from the
first synthetic division.
This was obtained from the
second synthetic division.
Set each factor equal to zero.
Solve.
42  4 1  6   4  8
x
2
2 1
4  2i 2

 2  i 2
2
4 

The solution set of the original equation is 2,  2  i 2 ,  2  i 2

Properties of Polynomial Equations
As we noticed from Examples 4 and 5:
1. If a polynomial equation is of degree n, then counting multiple roots
separately, the equation has n roots.
2.
If a + bi is a root of a polynomial equation with real coefficients
(b  0), then the complex number a - bi is also a root.
Complex imaginary roots, if they exist, occur in conjugate pairs.
The Fundamental Theorem of
Algebra
Remember that roots of 3,  2, etc. are complex roots:
3 can be written 3  0i, and  2 can be written as  2  0i.
The Linear Factorization
Theorem
Example 6
(# 28)
Find a 3rd degree polynomial function f ( x )
with real coefficients that has 6 and  5  2i
as zeros, and such that f (2)  636
Descartes’s Rule of Signs
Let f (x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial
with real coefficients.
1. The number of positive real zeros of f is either equal to the number
of sign changes of f (x) or is less than that number by an even
integer.
If there is only one variation in sign, there is exactly
one positive real zero.
2. The number of negative real zeros of f is either equal to the number
of sign changes of f (-x) or is less than that number by an even
integer. If f (-x) has only one variation in sign, then f has exactly one
negative real zero.
Descartes’s Rule of Signs
Determine the possible numbers of positive and negative
real zeros of f ( x) = x 3 + 2x 2 - 5x - 6.
1) To find possibilities for positive real zeros, count the number
of sign changes in the equation for f ( x). There is one variation
in sign change f ( x) = x 3 + 2x 2 - 5x - 6, so there is one positive real zero.
2) Now, substitute in - x for x:
( )
2
f (-x) = (-x) + 2 -x - 5(-x) - 6
3
f (-x) = - x 3 + 2x 2 + 5x - 6
There are two sign changes so there are either 2 negative
real zeros or none. There has to be 2 to give you a total of 3.
The zeros are 2, -1, - 3.
Example 7
For f ( x)  4 x 4  x3  5 x 2  2 x  6
1) How many positive and negative zeros
are there?
(# 38)
2) Find all Possible Rational Zeros.
3) Use a graphing utility to find the first
real zero of the function.
Confirm using synthetic division.
4) Find all remaining zeros of the function.
Practice 1
List all possible rational zeros of the
function f ( x)  x3  3x 2  6 x  8
1

,  1,  2,  4,  8
(a)
2
(b) 1,  2,  4,  8
(c)  1 ,  2,  4
2
(d) 1
Practice 2
Find a third-degree polynomial
function f ( x ) with real coefficients
that have 1 and 2i as zeros and such
that f (1)  0 .
(a) f ( x)  x 3  x 2  4 x  4
(b) f ( x)  2 x 3  x 2  4 x  8
(c) f ( x)  2 x 3  2 x 2  8 x  8
(d) f ( x)  x 3  x 2  4 x  4
Practice 3
What are the zeros of the function
f ( x)  x3  2 x2  8 x 16 ?
Find the first zero using a graphing utility.
(a) 2,  2i 2, 2i 2
2,  2 2, 2 2
(c) 2,  2i, 2i
(d) 2,  2, 2
(b)