Chapter 5.5
Special Cases of Factoring
Difference of Two Squares
1. Check to see if there is a GCF.
2. Write each term as a square.
3. Write those values that are squared as the product of a
sum and a difference.
a2 – b2 = (a + b)(a – b)
1. Factor.
64x2 – 1
( 8x )2 – ( 1 )2
1. GCF = 1
2. Write as squares
3. (sum)(difference)
(8x + 1)(8x – 1)
2. Factor.
36x2 – 49
( 6x )2 – ( 7 )2
1. GCF = 1
2. Write as squares
3. (sum)(difference)
(6x + 7)(6x – 7)
3. Factor.
100x2 – 81y2
(10x )2 – (9y)2
1. GCF = 1
2. Write as squares
3. (sum)(difference)
(10x + 9y)(10x – 9y)
4. Factor.
x8 – 1
2
4
(x )
– (1)
1. GCF = 1
2
(x4 + 1)(x4 – 1)
2. Write as squares
3. (sum)(difference)
( x2 )2– (1)2
(x4 + 1)(x2 + 1)(x2 – 1)
(x4 + 1)(x2 + 1)(x + 1)(x – 1)
Perfect Square Trinomials
1. Check to see if there is a GCF.
2. Determine if the 1st and 3rd terms are perfect squares.
3. Determine if the 2nd term is double the product of the
values whose squares are the 1st and 3rd terms.
4. Write as a sum or difference squared.
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
5. Factor.
x2 + 10x + 25
(x + 5)2
GCF = 1
2. Are the 1st and 3rd
terms perfect squares
x2 = (x)2
25 =
(5)2
√
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(x)(5) = 10x
√
4. Write as a sum squared.
6. Factor.
25x2 – 30x + 9
(5x – 3)2
GCF = 1
2. Are the 1st and 3rd
terms perfect squares
25x2 = (5x)2
9=
Used -3 because the
second term is – 30x
(-3)2
√
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(5x)(-3) = -30x √
4. Write as a difference squared.
7a. Factor.
25x2 + 60xy + 36y2
(5x + 6y)2
GCF = 1
2. Are the 1st and 3rd
terms perfect squares
25x2 = (5x)2
36y2
=
(6y)2
√
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(5x)(6y) = 60xy
√
4. Write as a sum squared.
7b. Factor.
64x6 – 48x3 + 9
(5x – 3)2
Used -3 because the
second term is – 48x3
GCF = 1
2. Are the 1st and 3rd
terms perfect squares
2
64x6 = (8x3)
√
2
9 = (-3)
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(8x3)(-3) = -48x3 √
4. Write as a difference squared.
8. Factor.
9x2 + 15x + 4
Not a perfect square trinomial
Use trial and error
or
the grouping method
GCF = 1
2. Are the 1st and 3rd
terms perfect squares
9x2 = (3x)2
4=
(2)2
√
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(3x)(2) = 12x
12x ≠ 15x
8. Factor.
9x2 + 15x + 4
1. GCF = 1
2. Grouping Number.
9x2 + 3x + 12x + 4
(9)(4) = 36
3. Find 2 integers whose
product is 36 and sum is 15.
1, 36
2, 18
3, 12
4. Split into 2 terms.
8. Factor.
9x2 + 15x + 4
5. Factor by grouping.
GCF = 3x
9x2 + 3x + 12x + 4
4
3x
3x( 3x + 1) + 4(3x + 1 )
(3x + 1)
(3x + 1)
(3x + 1)( 3x + 4)
GCF = 4
GCF = (3x + 1)
9. Factor.
20x2 – 45
5(4x2 – 9)
( 2x )2 – ( 3 )2
5(2x + 3)(2x – 3)
1. GCF = 5
2. Write as squares
3. (sum)(difference)
10. Factor.
75x2 – 60x + 12
3( 25x2
– 20x + 4)
3(5x – 2)2
Used -2 because the
second term is – 20x
GCF = 3
2. Are the 1st and 3rd
terms perfect squares
25x2 = (5x)2
4=
(-2)2
√
3. Is 2nd term double the product
of the values whose squares
are the 1st and 3rd terms
2(5x)(-2) = -20x √
4. Write as a difference squared.
Chapter 5.5
Special Cases of Factoring