§ 5.5
Factoring Special Forms
A Strategy for Factoring Polynomials,
page 363
1.
If there is a common factor, factor out the GCF or factor out a common factor with
a negative coefficient.
2.
Determine the number of terms in the polynomial and try factoring as follows:
(a) If there are two terms, can the binomial be factored by using one of the
following special forms.
2
2
Difference of two squares:
A  B   A  B  A  B 
///////////////////////////////////////////
Sum and Difference of two cubes:
(b) If there are three terms,
2
2
2
A  2 AB  B   A  B 
If is the trinomial a perfect square trinomial
2
2
2
A  2 AB  B   A  B 
use one of the adjacent forms:
If the trinomial is not a perfect square trinomial,
If a is equal to 1, use the trial and error
If a is > than 1, use the grouping method
(c) If there are four or more terms, try factoring by grouping.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 5.6
The Difference of Two Squares
The Difference of Two Squares
If A and B are real numbers, variables, or algebraic
expressions, then
A  B   A  B  A  B  .
2
2
In words: The difference of the squares of two terms,
factors as the product of a sum and a difference of those
terms.
P 353
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 5.5
The Difference of Two Squares
EXAMPLE
25 x  9 y .
4
Factor:
6
SOLUTION
We must express each term as the square of some monomial.
Then we use the formula for factoring A 2  B 2   A  B  A  B  .
25 x  9 y
4
6
5 x   3 y 
2 2
5 x
2
 3y
3 2
3
5 x
Express as the difference of
two squares
2
 3y
3

Factor using the Difference
of Two Squares method
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 5.5
The Difference of Two Squares
Check Point 1a
Factor
16 x  25
2
 4 x  5  4x
- 5
Check Point 1b
100 y  9 x
6
Factor
10 y
10 y
3
3
 3x
2
4
3x
10 y
3
2
 3x
2

P 354
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 5.2
The Difference of Two Squares
EXAMPLE
Factor:
6x  6 y .
2
2
SOLUTION
The GCF of the two terms of the polynomial is 6. We begin by
factoring out 6.
6x  6 y
2

6 x y
2
2
2

6  x  y  x  y 
Factor the GCF out of both
terms
Factor using the Difference
of Two Squares method
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 5.5
The Difference of Two Squares
Check Point 2
Factor
6 y  6x y
2
7
6 y (1  x y )
2
6y 1  xy
3
6
1  xy 
3
P 355
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 5.2
The Difference of Two Squares
EXAMPLE
Factor completely: x 4  1.
SOLUTION
 
x 1  x
4
2 2
1
2
Express as the difference of
two squares
  x  1 x  1
2

2

 x 1 x 1
2
2
2
The factors are the sum and
difference of the expressions
being squared

The factor x 2  1 is the
difference of two squares and
can be factored
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 5.5
The Difference of Two Squares
CONTINUED
  x  1 x  1 x  1
2
The factors of x  1 are the
sum and difference of the
expressions being squared
Thus, x 4  1  x 2  1 x  1 x  1 .
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 5.5
2
The Difference of Two Squares
Check Point 3
16 x  81
4
Factor
4x
4x
2
2
 9 4 x  9 
2
 9  2 x  3  2 x  3 
P 355
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 5.2
Factoring Completely
EXAMPLE
Factor completely:
x  3 x  9 x  27 .
3
2
SOLUTION
x  3 x  9 x  27
3
2
 x  3 x
3
 x
2
2
   9 x  27 
x  3  9x  3
  x  3  x  9 
2
  x  3  x  3  x  3 
Group terms with common
factors
Factor out the common factor
from each group
Factor out x + 3 from both
terms
Factor x 2  9 as the difference
of two squares
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 5.5
The Difference of Two Squares
Check Point 4
Factor
x  7 x  4 x  28
3
2
 x  7 x
2
4

 x  7  x  2  x  2 
P 355
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 5.2
Factoring Special Forms
Factoring Perfect Square Trinomials
Let A and B be real numbers, variables, or algebraic
expressions.
1) A  2 AB  B   A  B 
2
2
2
2 ) A  2 AB  B   A  B 
2
2
2
P 356
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 5.5
Factoring Perfect Square Trinomials
EXAMPLE
Factor:
16 x  40 xy  25 y .
2
2
SOLUTION
We suspect that 16 x 2  40 xy  25 y 2 is a perfect square trinomial
because 16 x 2  4 x 2 and 25 y 2   5 y 2 . The middle term can
be expressed as twice the product of 4x and -5y.
16 x  40 xy  25 y
2
2
 4 x   2  4 x    5 y    5 y 
2
 4 x  5 y 
2
2
Express in
Factor
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 5.5
A  2 AB  B
2
2
form
Factoring Perfect Square Trinomials
Check Point 5a
Factor
x  6x  9
2
x  3
2
Check Point 5b
Factor
16 x  40 xy  25 y
2
2
 4 x  5 y 2
Check Point 5b
Factor
4 y  20 y  25
4
2 y
2
2
5

2
P 357
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 5.2
Grouping & Difference of Two Squares
EXAMPLE, use for #65 and #67 (not on test)
Factor:
x  x  6 x  9.
4
2
SOLUTION
x  x  6x  9
4
2
 x  x  6 x  9 
4
2
 x  x  3
4
2
   x  3
 x
2 2
Group as x 4 minus a perfect
square trinomial to obtain a
difference of two squares
Factor the perfect square
trinomial
2
Rewrite as the difference of
two squares
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 5.5
Grouping & Difference of Two Squares
CONTINUED
  x   x  3  x   x  3 
2
2
  x  x  3  x  x  3 
2
2
Factor the difference of two
squares. The factors are the
sum and difference of the
expressions being squared.
Simplify
Thus, x 4  x 2  6 x  9  x 2  x  3 x 2  x  3  .
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 5.5
DONE
The Difference of Two Squares
Check Point 6
Factor
x  10 x  25  y
2
2
 x  5  y  x  5 - y 
Check Point 7
Factor
a  b  4b  4
2
2
 a  b  2  a  b  2 
P 357-8
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 5.2
Special Polynomials
In this section we will consider some polynomials that have special forms
that make it easy for us to see how they factor. You may look at a
polynomial and say, “Oh, that’s just a difference of squares” or “I think we
have a sum of cubes here.” When you have a special polynomial, in
particular one that is a difference of two squares, a perfect square
polynomial, or a sum or difference of cubes, you will have a factoring
formula memorized and will know how to proceed.
That’s why these polynomials are “special”. They may just become our best
friends among the polynomials.….
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 5.5
The Sum & Difference of Two Cubes
Factoring the Sum & Difference of Two Cubes
1)
Factoring the Sum of Two Cubes:
A  B   A  B  A  AB  B
3
3
Same Signs
2
2

2

Opposite Signs
2) Factoring the Difference of Two Cubes:
A  B   A  B  A  AB  B
3
3
Same Signs
2
Opposite Signs
Blitzer, Intermediate Algebra, 5e – Slide #21 Section 5.5
The Sum & Difference of Two Cubes
EXAMPLE
Factor: x 3 y 3  64 .
SOLUTION
x y  64
3
3
  xy   4
3
3

  xy  4   xy    xy 4  4
2
2
  xy  4  x y  4 xy  16 
2
2

Rewrite as the Sum of Two
Cubes
Factor the Sum of Two Cubes
Simplify
Thus, x 3 y 3  64   xy  4 x 2 y 2  4 xy  16  .
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 5.5
The Sum & Difference of Two Cubes
EXAMPLE
Factor: 125 x 6  64 y 6 .
SOLUTION
125 x  64 y
6

2 3

2
 5x
6
  4 y 
 5x  4 y
2 3
2
5 x   5 x 4 y   4 y 
2 2
2
2 2
2
Rewrite as the Difference of
Two Cubes
 Factor the Difference of Two
Cubes
 5 x  4 y
2
2
25 x
4
 20 x y  16 y
2
2
4

Simplify
Thus, 125 x 6  64 y 6  5 x 2  4 y 2 25 x 4  20 x 2 y 2  16 y 4  .
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 5.5