Download Chapter 3 Section 4 - Canton Local Schools

Chapter 3
Polynomial and
Rational
Functions
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All rights reserved
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1
SECTION 3.4
The Real Zeros of a Polynomial Function
OBJECTIVES
1
2
3
4
5
Review real zeros of polynomials.
Use the Rational Zeros Theorem.
Find the possible number of positive and
negative zeros of polynomials.
Find the bounds on the real zeros of
polynomials.
Learn a procedure for finding the real zeros
of a polynomial function.
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2
REAL ZEROS OF A POLYNOMIAL
FUNCTION
1. Definition If c is a real number in the
domain of a function f and f(c) = 0, then c is
called a real zero of f.
a. Geometrically, this means that the graph
of f has an x-intercept at x = c.
b. Algebraically, this means that (x – c) is
a factor of f(x).
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REAL ZEROS OF A POLYNOMIAL
FUNCTION
2. Number of real zeros A polynomial
function of degree n has, at most, n real
zeros.
3. Polynomial in factored form If a
polynomial F(x) can be written in factored
form, we find its zeros by solving F(x) = 0,
using the zero product property of real
numbers.
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REAL ZEROS OF A POLYNOMIAL
FUNCTION
4. Intermediate Value Theorem If a
polynomial F(x) cannot be factored, we find
approximate values of the real zeros (if any)
of F(x) by using the Intermediate
Value Theorem.
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THE RATIONAL ZEROS THEOREM
If F x   an x  an1 x
n
n1
 ...  a2 x  a1 x  a0
2
is a polynomial function with integer
p
coefficients (an ≠ 0, a0 ≠ 0) and
is a
q
rational number in lowest terms that is a
zero of F(x), then
1. p is a factor of the constant term a0;
2. q is a factor of the leading coefficient an.
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EXAMPLE 1
Using the Rational Zeros Theorem
Find all the rational zeros of
3
2
F  x   2 x  5 x  4 x  3.
Solution
List all possible zeros
p
Factors of the constant term,  3

q Factors of the leading coefficient, 2
Factors of  3 :  1,  3
Factors of 2 :
 1,  2
1
3
Possible rational zeros are:  1,  ,  ,  3.
2
2
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EXAMPLE 1
Using the Rational Zeros Theorem
Solution continued
Begin testing with 1. if it is not a rational zero,
then try another possible zero.
1 2 5 4 3
2 7
3
2 7 3 0
The remainder of 0 tells us that (x – 1) is a
factor of F(x). The other factor is 2x2 + 7x + 3.
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EXAMPLE 1
Using the Rational Zeros Theorem
Solution continued
1


The solution set is 1,  ,  3 .
2


1
The rational zeros of F are 3,  , and 1.
2
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EXAMPLE 2
Solving a Polynomial Equation
Solve 3x3 – 8x2 – 8x + 8 = 0.
Solution
Possible
rational
roots
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EXAMPLE 2
Solving a Polynomial Equation
Solution continued
From the graph, it appears
that an x-intercept is
between 0.5 and 1.
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EXAMPLE 2
Solving a Polynomial Equation
Solution continued
2
We check whether x  is a root by synthetic
3
division.
2

 x   is a factor of the original equation with
3

the depressed equation 3x2 – 6x – 12 = 0.
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EXAMPLE 2
Solving a Polynomial Equation
Solution continued
So
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EXAMPLE 2
Solving a Polynomial Equation
Solution continued
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EXAMPLE 2
Solving a Polynomial Equation
Solution continued
You can see from the graph that x  5 are also
the x-intercepts of the graph.
The three real roots of the given equation are
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DESCARTE’S RULE OF SIGNS
Let F(x) be a polynomial function with real
coefficients and with terms written in
descending order.
1. The number of positive zeros of F is
either equal to the number of variations of
sign of F(x) or less than that number by
an even integer.
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DESCARTE’S RULE OF SIGNS
2. The number of negative zeros of F is
either equal to the number of variations of
sign of F(–x) or less than that number by
an even integer.
When using Descarte’s Rule, a zero of
multiplicity m should be counted as m zeros.
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EXAMPLE 3
Using Descarte’s Rule of Signs
Find the possible number of positive and
negative zeros of
5
3
2
f x   x  3x  5x  9x  7.
Solution
There are three variations in sign in f(x).
f x   x  3x  5x  9x  7
5
3
2
The number of positive zeros is either 3
or (3 – 2 =) 1.
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EXAMPLE 3
Using Descarte’s Rule of Signs
Solution continued
There are two variations in sign in f(–x).
f x   x   3x   5 x   9 x   7
5
3
2
 x 5  3x 3  5x 2  9x  7
The number of negative zeros is either 2
or (2 – 2 =) 0.
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RULES FOR BOUNDS
Let F(x) be a polynomial function with real
coefficients and a positive leading coefficient.
Suppose F(x) is synthetically divided by x – k.
Then
1. If k > 0, and each number in the last row is
either zero or positive, then k is an upper
bound on the zeros of F(x).
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RULES FOR BOUNDS
2. If k < 0, and numbers in the last row
alternate in sign, then k is a lower bound on
the zeros of F(x).
Zeros in the last row can be regarded as
positive or negative.
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EXAMPLE 4
Finding the Bounds on the Zeros
Find upper and lower bounds on the zeros of
4
3
2
F x   x  x  5x  x  6.
Solution
The possible zeros are: ±1, ±2, ±3, and ±6.
There is only one variation in sign so there is
one positive zero. Use synthetic division with
positive numbers until last row is all positive
or 0. This first
3 1 1 5 1 6
occurs for 3.
3 6 3 6
1 2 1 2 0
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EXAMPLE 4
Finding the Bounds on the Zeros
Solution continued
So 3 is upper bound.
Use synthetic division with negative numbers
until last row alternates in sign. This first
occurs for –2.
2 1 1 5 1 6
2 6 2 6
1 3 1 3 0
So –2 is lower bound.
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FINDING THE REAL ZEROS OF A
POLYNOMIAL
Step 1 Find the maximum number of real
zeros by using the degree of the polynomial
function.
Step 2 Find the possible number of positive
and negative zeros by using Descartes’s Rule
of Signs.
Step 3 Write the set of possible rational zeros.
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FINDING THE REAL ZEROS OF A
POLYNOMIAL
Step 4 Test the smallest positive integer in the
set in Step 3, the next larger, and so on, until
an integer zero or an upper bound of the zeros
is found.
a. If a zero is found, use the function
represented by the depressed equation in
further calculations.
b. If an upper bound is found, discard all
larger numbers in the set of possible
rational zeros.
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FINDING THE REAL ZEROS OF A
POLYNOMIAL
Step 5 Test the positive fractions that remain
in the set in Step 3 after considering any
bound that has been found.
Step 6 Use modified Steps 4 and 5 for
negative numbers in the set in Step 3.
Step 7 If a depressed equation is quadratic,
use any method (including the quadratic
formula) to solve this equation.
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Find the real zeros of
Solution
1. f has, at most, 5 real zeros
2.
There are three variations of sign in f(x).
So f has 3 or 1 positive zeros.
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
2. continued
There are two variations of signs in f(–x).
So f has 2 or 0 negative zeros.
3. Possible rational zeros are
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
4. Test for zeros, 1, 2, 3, 4, 6, and 12, until a
zero or an upper bound is found.
(x – 2) is a factor of f, so
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
4. continued Consider
The possible positive rational zeros of Q1(x)
are
Discard 1 since it is not a zero of the original
function.
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
4. continued Test 2:
2 is not a zero, but is an upper bound.
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
1
3
5. Test remaining positive entries and .
2
2
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
1
5. continued Since is a zero,
2
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
6. Find the negative zeros of
Q2(x) = x3 + 2x2 – 3x – 6.
Test –1, –2, –3, and –6.
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EXAMPLE 5
Finding the Real Zeros of a Polynomial
Function
Solution continued
7. We can solve the depressed equation
x2 – 3 = 0 to obtain x   3. The real zeros of f
1
are 2,  3, , 3, and 2. Because f(x) has, at
2
most, five real zeros, we have found all of them.
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