The Factor

1
Chapter 4 Quadratic and Polynomial Equations
4.9A
The Factor Theorem
4.9A.1
MATHPOWERTM 20, WESTERN EDITION
2
The Factor Theorem
According to the remainder theorem, if a number a
is substituted for x in a polynomial, the value
obtained is the remainder when the polynomial is
divided by x - a. If this remainder is 0, then x
- a is a factor of the polynomial. This special
case of the remainder theorem is called the
factor theorem.
Factor Theorem
Factor Theorem
Factor Theorem
If x a is substituted into a polynomial in x,
and the resulting value is 0, then x - a is a
factor of the polynomial.
4.9A.2
3
Applying the Remainder Theorem
Determine if the following binomials are factors
of f(x) x3 - 6x2 3x 10.
a) x - 2
b) x 1
f(-1) (-1)3 - 6 (-1)2 3 (-1) 10
-1 - 6 - 3 10 0
f(2) (2)3- 6(2)2 3(2) 10 8 - 24
6 10 0
Since f(-1) 0, x 1 is a factor.
Since f(2) 0, x - 2 is a factor.
c) x - 5
Since f(5) 0, x - 5 is a factor.
f(5) (5)3- 6(5)2 3(5) 10 125 -
150 15 10 0
4.9A.3
4
The Factor Theorem
Note f(x) x3 - 6x2 3x 10 has three
factors f(x) (x - 5)(x
1)(x - 2)
The product of the constant terms is (-5)(1)(-2)
10. 10 is also the constant term of the
polynomial.
If a polynomial has any factor in the form x -
a, then a is a factor of the constant term of
the polynomial.
If x - a is a factor of the polynomial, f(a) 0,
then we say that a is a zero of the polynomial.
4.9A.4
5
The Integral Zero Theorem
If x a is an integral zero of a polynomial,
with integral coefficients, then a is a factor
of the constant term of the polynomial.
Therefore, use the Integral Zero Theorem, by
first finding factors of the constant term to
help in finding factors of the polynomial.
4.9A.5
6
Applying the Factor Theorem
Factor P(x) x3 2x2 - 5x - 6.
1. Find a value of x so that P(x) 0.
Potential Zeros 1, 2, 3, 6
2. Using substitution, use the potential
zeros to find a zero of the polynomial.
Try P(1) P(1) (1)3 2(1)2 - 5(1) - 6
8
Try P(-1) P(-1) (-1)3 2(-1)2 - 5(-1) -
6 P(-1) 0
According to the integral zero theorem, the
numbers to try are factors of 6. Therefore the
Potential Zeros are 1, 2, 3, 6
Since P(-1) 0, then x 1 is a factor of the
polynomial.
Since P(1) ? 0, then x - 1 is not a factor. Try
another potential zero.
1 1 2 -5 -6
3. Once you have found a factor of the
polynomial, use synthetic division to
find the remaining factors.
1
1
-6
1
1
-6
0
Factor x2 x - 6 (x 3)(x - 2)
4. Factor the remaining binomial.
Therefore, P(x) (x 1)(x 3)(x - 2).
4.9A.6
7
Assignment
Suggested Questions
Pages 209 and 210 5, 6, 24, 43, 45, 55-63 odd, 78
4.9A.7
8
Chapter 4 Quadratic and Polynomial Equations
4.9B
Graphs of a Polynomial
4.9B.8
MATHPOWERTM 11, WESTERN EDITION
9
Factoring a Polynomial
Factor Completely f(x) x3 - 9x2 23x - 15
Potential Zeros 1, 3, 5, 15
f(1) 0
Thus, (x - 1) is a factor.
-1 1 -9 23 -15 -1
8 -15 1 -8 15 0
f(x) (x - 1)(x2 - 8x 15) f(x) (x - 1)(x -
5)(x - 3)
4.9B.9
10
Summary of the Graphs of Polynomials.

• The graphs of odd-degree polynomials act like a
  line.
• The arrows will be pointing in opposite
  directions.
• If the leading coefficient is positive, then the
  graph
• starts down and finishes up.
• If the leading coefficient is negative, then the
  graph
• starts up and finishes down.
• Keep in mind that the x-intercepts of the graphs
  are
• the zeros of the polynomial function.
• The degree of the polynomial can be determined
  by
• the number of changes of direction of the
  graph.
• The number of zeros will also determine the
  degree of
• the polynomial.

4.9B.10
11
Understanding The Graphs of Polynomials
1
2
2
3
3
1
Degree 3 Positive
Degree 3 Negative
4.9B.11
12
Understanding The Graphs of Polynomials
A Degree 3 Negative
Has one zero at x -2, and the other two zeros
at x 2. We say that the zero at x 2 has a
multiplicity of two.
Notice that the graph bounces off the x-axis
when there is a multiplicity of two.
4.9B.12
13
The Degree of a Polynomial
For an even-degree polynomial

• The two arrows point in the same direction.
• A positive even-degree polynomial will start up
• and finish up.
• A negative even-degree polynomial will start
  down
• and finish down.

4.9B.13
14
The Graphs of Even-Degree Polynomials
Positive and Even
Negative and Even
4.9B.14
15
The Graphs of Even-Degree Polynomials
Degree 4 Polynomial
Zeros are x -2 and x 2, each with
a multiplicity of two.
4.9B.15
16
Assignment
Suggested Questions Pages 212 and 213
4.8B.16
17
Chapter 4 Quadratic and Polynomial Equations
4.9C
Rational Zero Theorem
4.9C.17
MATHPOWERTM 11, WESTERN EDITION
18
Rational Zero Theorem
The factor theorem can be extended to
include polynomials in which the leading
coefficient of the highest degree term is not 1.
The zeros of the function f(x) (2x - 3)(5x -
7)(x 11) are
Two of the
zeros are rational zeros and one is an integral
zero. Note that if we were to expand this
polynomial, the leading coefficient would be
greater than 1 f(x) 10x3 81x2 - 298x 231
Rational Zero Theorem If x
is a rational zero of a polynomial
with integral coefficients, then a is a factor of
the constant term of the polynomial and b is a
factor of the coefficient of the highest-degree
term in the polynomial.
4.9C.18
19
Rational Zero Theorem contd
Given the polynomial f(x) 10x3 81x2 - 298x
231, the zeros are
Note the numerators of the zeros
3, 7, 11
They are all factors of the constant term 231.
Note the denominators of the zeros
2, 5, 1
They are all factors of the leading coefficient.
4.9C.19
20
Applying the Rational Zero Theorem
All rational zeros of an integral polynomial are
of the form
, where the numerator a is a factor
of the constant term and the denominator b is a
factor of the leading coefficient.
Find the potential zeros of P(x) 6x3 - 6x - 21.
Factors of the constant term are factors of 21
Factors of the leading coefficient are factors
of 6
b
1, 2, 3, 6
a
1, 3, 7, 21
Potential Zeros are in the form of
Therefore the potential zeros are
4.9B.20
21
Factoring Polynomials
Factor completely the polynomial P(x) 6x3
13x2 x - 2.
List potential zeros
Factors of 2 1, 2
Factors of 6 1, 2, 3, 6
Potential Zeros are
It is easier to test the integral values first
2 6 13 1 -2
P(1) 6(1)3 13(1)2 (1) - 2 18
2
-2
12
(Not a zero)
6
0
1
-1
P(-2) 6(-2)3 13(-2)2 (-2) - 2 0
Factor 6x2 x - 1
Since P(-2) 0, x 2 is a factor.
P(x) (x 2)(3x - 1)(2x 1)
4.9C.21
22
Assignment
Suggested Questions
Pages 210 and 211 66, 68, 71, 74, 79, 84, 85, 86,
89 Page 212 1 All
4.9B.22