Factoring Polynomials

1
Factoring Polynomials

• Chapter 6

2
Chapter Sections
6.1 Greatest Common Factor and Factoring by
Grouping 6.2 Factoring Trinomials of the Form
x2 bx c 6.3 Factoring Trinomials of the
Form ax2 bx c, a ? 1 6.4 Factoring Special
Products 6.5 Summary of Factoring
Techniques 6.6 Solving Polynomial Equations by
Factoring 6.7 Modeling and Solving Problems
with Quadratic Equations
3
Greatest Common Factor and Factoring by Grouping

• Section 6.1

4
Factors
4 9 36 3(x 2) 3x 6 (2x 7)(3x 5)
6x2 11x 35
The expressions on the left side are called
factors of the expression on the right side.
The greatest common factor (GCF) of a list of
algebraic expressions is the largest expression
that divides evenly into all the expressions.
5
Greatest Common Factor
Finding the Greatest Common Factor Step 1 Write
each number as a product of prime factors. Step
2 Determine the common prime factors. Step 3
Find the product of the common factors found in
Step 2. This number is the GCF.
Example Find the GCF of 16 and 20.
16 2 2 2 2
20 2 2 5
The GCF is 2 2 4.
6
Greatest Common Factor
Example Find the GCF of 60, 75, and 135.
60 2 2 3 5
75 3 5 5
135 3 3 3 5
The GCF is 3 5 15.
7
The GCF as a Binomial
Example Find the GCF of 6(x y) and 15(x y)3.
6(x y) 2 3 (x y)
15(x y)3 3 5 (x y) (x y) (x
y)
The GCF is 3 (x y) 3(x y).
8
Finding the GCF
Steps to Find the Greatest Common Factor of Two
or More Expressions Step 1 Find the GCF of the
coefficients of each variable expression. Step 2
For each variable expression, determine the
smallest exponent that the variable expression is
raised to. Step 3 Find the product of the
common factors found in Steps 1 and 2. This
expression is the GCF.
9
Factoring Polynomials
Steps to Factor a Polynomial Using the GCF Step
1 Identify the GCF of the terms that make up
the polynomial. Step 2 Rewrite each term as the
product of the GCF and the remaining factor. Step
3 Use the Distributive Property in reverse to
factor out the GCF. Step 4 Check Use the
Distributive Property to verify that the
factorization is correct.
10
Factoring Polynomials
Example Factor the trinomial 36a6 45a4 18a2
by factoring out the GCF.
Step 1 Find the GCF.
GCF 9a2
Step 2 Rewrite each term as the product of the
GCF and the remaining term.
36a6 45a4 18a2 9a2 4a4 9a2 5a2 9a2
2
Step 3 Factor out the GCF.
36a6 45a4 18a2 9a2(4a4 5a2 2)
Step 4 Check.
9a2(4a4 5a2 2) 36a6 45a4 18a2
?
11
Factoring Out a Negative Number
Example Factor 3x6 9x4 18x by factoring
out the GCF
Step 1 Find the GCF.
GCF 3x
Step 2 Rewrite each term as the product of the
GCF and the remaining term.
3x6 9x4 18x 3x x5 ( 3x)( 3x3)
( 3x) 6
Step 3 Factor out the GCF.
3x6 9x4 18x 3x(x5 3x3 6)
Step 4 Check.
3x(x5 3x3 6) 3x6 9x4 18x
?
12
Factoring Out a Binomial
Example Factor out the greatest common binomial
factor 6(3x y) z(3x y)
6(3x y) z(3x y) (3x y)(6 z)
Check
(3x y)(6 z) 6(3x y) z(3x y)
?
13
Factoring by Grouping
Steps to Factor by Grouping Step 1 Group the
terms with common factors. Step 2 In each
grouping, factor out the greatest common
factor. Step 3 Factor out the common factor
that remains. Step 4 Check your work.
14
Factoring by Grouping
Example Factor by grouping x2 7x 3x 21
x2 7x 3x 21
x2 7x 3x 21 x(x 7)
3(x 7)
(x 3)(x 7)
Factor out the common factor x 7.
(x 3)(x 7) x2 7x 3x 21
Check
?
15
Factoring by Grouping
Example Factor by grouping xy 4x 3y 12
Step 1 Group the terms with the common factors.
(xy 4x) ( 3y 12)
Step 2 Factor out the common factor in each
group.
xy 4x 3y 12 x(y 4) ( 3)(y 4)
Step 3 Factor out the common factor that
remains.
(x 3)(y 4)
xy 4x 3y 12 (x 3)(y 4)
Step 4 Check.
(x 3)(y 4) xy 4x 3y 12
?
16
Factoring Trinomials of the Form x2 bx c

• Section 6.2

17
Quadratic Trinomials
A quadratic trinomial is a polynomial of the form
ax2 bx c, a ? 0 where a represents
the coefficient of the squared (second degree)
term, b represents the coefficient of the linear
(first degree) term and c represents the
constant.
3x2 4x 7
8a2 24a 10
6c2 c 25
When the trinomial is written in standard form
(or descending order of degree), the coefficient
of the squared term is called the leading
coefficient.
18
Trinomials of the Form x2 bx c
Factoring a Trinomial of the Form x2 bx
c Step 1 Find the pair of integers whose
product is c and whose sum is b. That is,
determine m and n such that mn c and m n
b. Step 2 Write x2 bx c (x m)(x
n). Step 3 Check your work by multiplying the
binomials using the FOIL method.
The coefficient of x is the sum of the two
numbers.
x2 11x 24 (x 3)(x 8)
The last term is the
product of the two numbers.
19
Trinomials of the Form x2 bx c
Example Factor x2 8x 15
x2 8x 15 (x ?)(x ?)
3
5
___ ___ 15
Find two numbers that we can multiply together to
get 15 and add together to get 8.
3
5
___ ___ 8
x2 8x 15 (x 3)(x 5)
Check
?
(x 3)(x 5) x2 8x 15
20
Trinomials of the Form x2 bx c, c lt 0
Example Factor x2 x 42
x2 x 42
(x ?)(x ?)
One factor will be positive and one will be
negative.
(? 6)
7
____ ____ ? 42
Find two numbers that we can multiply together to
get 42 and add together to get 1.
(? 6)
7
____ ____ 1
x2 x 42 (x 7)(x ? 6)
Check
?
(x 7)(x 6 ) x2 x 42
21
Trinomials of the Form x2 bx c
The following table summarizes the four forms for
factoring a quadratic trinomial in the form x2
bx c.
22
Prime Polynomials
A polynomial that cannot be written as the
product of two other polynomials (other than 1 or
1) is said to be a prime polynomial.
Example Factor 5x2 ? x ? 2
The Factors of 5 are
The Factors of 2 are
1 and 5
1 and 2
Possible Factors
Middle Term
(x 1)(5x 2)
3x
9x
(x 2)(5x 1)
The polynomial 5x2 ? x ? 2 is prime.
23
Trinomials with a Common Factor
Example Factor 2x2 32x 96
2x2 32x 96
2(x2 16x 48)
The common factor of 2 can be factored out.
x2 16x 48
(x ?)(x ?)
(? 4)
(? 12)
____ ____ 48
Find two numbers that we can multiply together to
get 48 and add together to get 16.
(? 4)
(? 12)
____ ____ 16
2x2 32x 96 2(x 12)(x ? 4)
Check
2(x 12)(x ? 4) 2(x2 16x 48)
?
2x2 32x 96
24
Negative Leading Coefficients
Example Factor x2 12x 36
It is easier to factor out a GCF of 1.
x2 12x 36 1(x2 12x 36)
1(x2 12x 36) 1(x ?)(x ?)
6
6
___ ___ 36
Find two numbers that we can multiply together to
get 36 and add together to get 12.
6
6
___ ___ 12
1(x2 12x 36) 1(x 6)(x 6)
(x 6)2
Check
(x 6)2 1(x 6)(x 6)
?
x2 12x 36
25
Factoring Trinomials of the Form
ax2 bx c, a ? 1

• Section 6.3

26
Trinomials of the Form ax2 bx c, a ? 1
When factoring trinomials of the form ax2 bx
c where a is not equal to 1, there are two
methods that can be used 1. Trial and error
2. Factoring by grouping
Factoring ax2 bx c, a ? 1 Using Trial and
Error a, b, and c Have No Common Factors Step 1
List the possibilities for the first terms of
each binomial whose product is ax2. Step 2 List
the possibilities for the last terms of each
binomial whose product is c. Step 3 Write out
all the combinations of factors found in Steps 1
and 2. Multiply the binomials out until a
product is found that equals the trinomial.
( __x )( __x ) ax2 bx c
27
The Trial and Error Method
Example Factor by trial and error 3x2 4x 1
The Factors of 3 are
The Factors of 1 are
1 and 3
1 and 1
Possible Factors
Middle Term
(x 1)(3x 1)
4x
(x 1)(3x 1) 3x2 4x 1
Check
?
28
The Trial and Error Method
Example Factor 5x2 2x ? 7
The sign of one factor will be positive, the sign
of the other factor will be negative.
The Factors of 5 Are
The Factors of 7 Are
1 and 5
1 and 7
Possible Factors
Middle Term
(x 1)(5x ? 7)
2x
2x
(x ? 1)(5x 7)
5x2 2x ? 7 (x 1)(5x 7)
(x 1)(5x 7) 5x2 2x ? 7
Check
?
29
The Trial and Error Method
Example Factor by trial and error 9x2 ? 13x
4
The Factors of 9 are
The Factors of 4 are
1 and 9
1 and 4
2 and 2
3 and 3
Possible Factors
Middle Term
(x 1)(9x 4)
13x
20x
(x 2)(9x 2)
15x
(3x 1)(3x 4)
12x
(3x 2)(3x 2)
9x2 ? 13x 4 (x 1)(9x 4)
Check
?
30
Factoring by Grouping
Factoring ax2 bx c, a ? 1 by Grouping a, b,
and c Have No Common Factors Step 1 Find the
value of ac. Step 2 Find the pair of integers
whose product is ac and whose sum is b. (Find m
and n so that mn ac and m n b.) Step 3
Write ax2 bx c ax2 mx nx c. Step 4
Factor the expression in Step 3 by grouping. Step
5 Check Multiply out the factored form.
31
Factoring by Grouping
Example Factor by grouping 2x2 9x 4
The value of ac 8.
Find two numbers that we can multiply together to
get 8 and add together to get 9.
1
8
___ ___ 8
1
8
___ ___ 9
2x2 9x 4 2x2 x 8x 4
Rewrite 9x as x 8x.
x(2x 1) 4(2x 1)
Factor by grouping.
(2x 1)(x 4)
Factor out (2x 1).
(2x 1)(x 4)
2x2 8x 1x 4
Check
2x2 9x 4
?
32
Factoring by Grouping
Example Factor by grouping 8x2 10x ? 3
The value of ac is ? 24.
(? 2)
12
____ ____ ? 24
Find two numbers that we can multiply together to
get ? 24 and add together to get 10.
(? 2)
12
____ ____ 10
8x2 10x ? 3 8x2 12x ? 2x ? 3
Rewrite 10x as 12x (?2x).
4x(2x 3) ? 1(2x 3)
Factor by grouping.
(2x 3)(4x ? 1)
Factor out (2x 3).
(2x 3)(4x ? 1) 8x2 10x ? 3
?
Check
33
Factoring Special Products

• Section 6.4

34
Perfect Square Trinomials
Perfect Square Trinomials a2 2ab b2 (a
b)2 a2 2ab b2 (a ? b)2
In order for a polynomial to be a perfect
square trinomial, two conditions must be
satisfied 1. The first and last terms must
be perfect squares. 2. The middle term must
equal 2 or 2 times the product of the
expressions being squared in the first and last
term.
35
Perfect Square Trinomials
Example Factor the following
The first term, x2, and third term, 16, are
perfect squares.
a.) x2 8x 16
The middle term, 8x, is 2 times the product of x
and 4.
x2 8x 16 (x 4)2
b.) 9x4 ? 30x2z 25z2
9x4 ? 30x2z 25z2 (3x2 5z)2
36
The Difference of Two Squares
Difference of Two Squares a2 b2 (a b)(a ? b)
Example Factor the following
a.) 3x2 ? 27
3x2 ? 27 3(x2 9)
3 is a common factor.
3(x 3)(x 3)
b.) 4x2 ? 25y4
4x2 ? 25y4 (2x 5y2)(2x 5y2)
37
The Sum of Two Cubes
The Sum of Two Cubes a3 b3 (a b) (a2 ? ab
b2)
Example Factor 27x3 125
a 3x
b 5
27x3 125 (3x)3 53
(3x 5) (3x)2 ? 3x5 52
(3x 5)(9x2 ? 15x 25)
38
The Difference of Two Cubes
The Difference of Two Cubes a3 ? b3 (a ? b) (a2
ab b2)
Example Factor x3 ? 64
a x
b 4
x3 ? 64 x3 43
(x ? 4) (x2 4x 42)
(x ? 4) (x2 4x 16)
39
Summary of Factoring Techniques

• Section 6.5

40
Steps for Factoring
Steps for Factoring Step 1 Is there a greatest
common factor? If so, factor out the GCF. Step
2 Count the number of terms. Step 3 (a) 2
terms (Binomials)

• Is it the difference of two squares? If so,
• a2 b2 (a b)(a b)

• Is it the difference of two cubes? If so,
• a3 b3 (a b)(a2 ab b2)

• Is it the sum of two cubes? If so,
• a3 b3 (a b)(a2 ab b2)

Continued.
41
Steps for Factoring
Steps for Factoring (Continued) (b) 3 terms
(Trinomials)

• Is it a perfect square trinomial? If so,
• a2 2ab b2 (a b)2 or a2 2ab b2 (a
  b)2

• Is the coefficient of the square term 1? If so,
• x2 bx c (x m)(x n) where mn c and m
  n b

• Is the coefficient of the square term ? 1? If
  so,
• a. Use factoring by grouping
• b. Use trial and error

(c) 4 terms

• Use factoring by grouping

Step 4 Check each factor to determine if it can
be factored again.
Step 5 Check your work by multiplying out the
factored form.
42
Factoring Completely
Example Factor completely 6x2 ? 6x ? 36
6x2 ? 6x ? 36 6(x2 ? x ? 6)
Factor out the GCF, 6.
3 terms, the coefficient of x2 is 1.
6(x ? 3)( x 2)
Factor.
3(2) 6 and 3 2 1
Check
6(x ? 3)( x 2) 6(x2 x 6)
6x2 6x 36

?
43
Factoring Completely
Example Factor completely 1 ? 16x4
1 and 16x4 are both perfect squares so we have
the difference of two squares.
1 ? 16x4
12 ? (4x2)2
(1 ? 4x2)(1 4x2)
Factor. a 1, b 4x2
This is also a difference of two squares.
(1 ? 2x)(1 2x)(1 4x2)
Factor.
Check
(1 ? 2x)(1 2x)(1 4x2) (1 ? 4x2)(1 4x2)
1 ? 16x4
?

44
Factoring Completely
Example Factor completely 80w3 ? 10
80w3 ? 10 10(8w3 ? 1)
Factor out the GCF, 10.
8w3 and 1 are both perfect cubes, so we have the
difference of two cubes.
10(2w)3 ? 13
a 2w, b 1.
10(2w ? 1)(4w2 2w 1)
Factor.
Check
10(2w ? 1)(4w2 2w 1) 10(8w3 ? 1)

80w3 ? 10
?
45
Solving Polynomial Equations by Factoring

• Section 6.6

46
Zero-Product Property

• A polynomial equation is any equation that
  contains a polynomial expression. The degree of
  a polynomial equation is the degree of the
  polynomial expression in the equation.

3z 7 1
4k2 5 0
8x4 4x3 5 12
Degree 1
Degree 2
Degree 4
The Zero-Product Property If the product of two
factors is zero, then at least one of the factors
is 0. That is, if ab 0, then a 0 or b 0 or
both a and b are 0.
47
Zero Factor Property
Example Solve (x 5)(x 4) 0
Set each expression equal to 0.
x 5 0 or x 4 0
x 5
x ? 4
The solution set is 4, 5.
Check
(x 5)(x 4) 0
(x 5)(x 4) 0
(5 5)(5 4) 0
( 4 5)( 4 4) 0
0(9) 0
9(0) 0
0 0
0 0
?
?
48
Quadratic Equations
A quadratic equation is an equation equivalent to
one of the form ax2 bx c 0 where
a, b, and c are real numbers and a ? 0.
Steps for Solving a Quadratic Equation by
Factoring Step 1 Write the equation in standard
form, ax2 bx c 0. Step 2 Factor the
expression on the left hand side of the
equation. Step 3 Set each factor found in Step
2 equal to zero using the Zero-Product
Property Step 4 Solve each first-degree
equation for the variable. Step 5 Check Check
your answers by substituting into the original
equation.
49
Solving Quadratic Equations
Example Solve 6x 18x2 0
6x 18x2 0
Put the equation in standard form.
18x2 6x 0
Factor.
6x(3x 1) 0
Set each factor equal to zero.
6x 0 3x 1 0
x 0
3x ?1
Check
6x 18x2 0
6(0) 18(0)2 0
0 0
?
?
50
Solving Quadratic Equations
Example Solve 9x2 81
9x2 81 0
Rewrite the equation in standard form.
9(x2 9) 0
Factor out the common factor, 9.
9(x 3)(x 3) 0
Factor the quadratic equation.
9 0 x 3 0 x 3 0
Set each expression equal to 0.
x 3
Solve each equation.
x ? 3
The solution set is 3, 3.
Check
9x2 81
9x2 81
9(? 3)2 81
9(3)2 81
9(9) 81
9(9) 81
?
?
51
Solving Quadratic Equations
Example Solve
Multiply each side by the LCD.
Simplify.
Remove parentheses.
Write in standard form.
Factor.
Continued.
52
Solving Quadratic Equations
Example continued
Set each factor equal to zero.
Solve for x.
Check
?
?
53
Solving Quadratic Equations
Example Solve 3x3 x2 14x
3x3 x2 14x 0
Rewrite the equation in standard form.
x(3x2 x 14) 0
Factor out the common factor, x.
x(x 2)(3x 7) 0
Factor the quadratic equation.
x 0 x 2 0 3x 7 0
Set each expression equal to 0.
Solve each equation.
x 0
x 2
54
Modeling and Solving Problems with Quadratic
Equations

• Section 6.7

55
Problems Involving Quadratic Equations

• Example
• The area of a rectangle is 84 square inches.
  Determine the length and width if the length is 2
  inches less than twice the width.

Step 1. Identify This is a geometry problem
involving the area of a rectangle.
2w 2
l
Step 2. Name
Area 84 l w
w
Let w width
Let l length
Step 3. Translate Because the length is 2
inches less than twice the width, we know that l
2w 2.
84 l w
84 (2w 2)w
Substitute.
Continued.
56
Problems Involving Quadratic Equations
Example continued
Step 4 Solve
84 (2w 2)w
84 2w2 2w
Distribute.
2w2 2w 84 0
Rewrite the equation in standard form.
2(w2 w 42) 0
Factor out the common factor, 2.
2(w 7)(w 6) 0
Factor the quadratic equation.
2 0 w 7 0 w 6 0
Set each expression equal to 0.
w 7 w 6
Solve the equations.
Continued.
57
Problems Involving Quadratic Equations
Example continued
The width of the rectangle is 7 inches.
The length is 2w 2 2(7) 2 12 inches.
Step 5 Check.
Area l w
7 12
84
?
The length is 2 inches less than twice the width.
12 2(7) 2
12 12
?
58
The Pythagorean Theorem
A right triangle is one that contains a right
angle, that is, an angle of 90. The side of the
angle opposite the 90 angle is called the
hypotenuse the remaining two sides are called
legs.
Pythagorean Theorem In a right triangle, the
square of the length of the hypotenuse is equal
to the sum of the squares of the lengths of the
legs. c2 a2 b2 or leg2 leg2
hypotenuse2
59
The Pythagorean Theorem

• Example
• Find the length of each leg of the right triangle.

The Pythagorean Theorem is used. The lengths of
the legs are 4x and 7x 1, and the length of the
hypotenuse is 9x 1.
a2 b2 c2
(4x)2 (7x 1)2 (9x 1)2
16x2 49x2 14x 1 81x2 18x 1
Continued.
60
The Pythagorean Theorem
Example continued
16x2 49x2 14x 1 81x2 18x 1
16x2 32x 0
Write in standard form.
16x(x 2) 0
Factor out the GCF.
16x 0
Set each factor equal to 0.
x 2 0
x 0
Solve each equation.
x 2
The length of one leg is 4(2) 8.
The length of the other leg is 7(2) 1 15.
The length of the hypotenuse is 9(2) 1 17.
Check 82 152 172
64 225 289
?
61
The Pythagorean Theorem

• Example
• A baseball diamond is square. Each side of the
  square is 90 feet long. How far is it from home
  plate to second base?

Step 1 Identify We want to know how far it is
from home plate to second base.
Home plate
Let c be the distance from home plate to second
base.
Step 2 Name
c
Second base
Continued.
62
The Pythagorean Theorem
Example continued
Step 3 Translate
c2 a2 b2
Use the Pythagorean Theorem.
c2 902 902
Substitute.
c2 8100 8100
Step 4 Solve
c2 16200
Continued.
63
The Pythagorean Theorem
Example continued
?127.3 is not used because length is never
negative.
Step 5 Check
c2 a2 b2
127.32 902 902
16205.29 8100 8100
16205.29 ? 16200
?
Step 6 Answer The distance from home plate to
second base is approximately 127.3 feet.