Worksheet___________________________________________________Name______
FACTORING AND SPECIAL PRODUCTS
There are certain types of products that must be memorized.

Difference of two squares
The square of the first term minus the square of the second term is equal to the product of the
sum and difference of the two terms.
(first term)2 – (second term)2 = (first term + second term)(first term – second term)
a2 – b2 = (a + b)(a – b)
a 2  a and
Note that
b2  b .
If the original problem does not have an explicit base with an exponent of 2, use the positive
square root of the expression to find the first and second terms. For example if 4x 2 is given, the
positive square root of 4x2, ( 4x 2 ) is 2x, so that we write (2x)2.
Example 1
Factor each polynomial.
(a) 9x2 – 25
(b) 16x4 – 81
(c) 2v2 - 72u2
Solution:
(a)
a
b
25 =5
9x 2 = 3x
(b)
a
16 x  4 x
4
2
b
81  9
9x2 – 25
= (3x)2 – (5)2 = (3x – 5)(3x + 5)
16x4 – 81
= (4x2)2 – (9)2 = (4x2 + 9)(4x2 – 9)
But 4x2 – 9 is also difference of two squares so that
a
4x 2 = 2x
Substituting we have,
(c)
4x2 – 9 = (2x)2 – (3)2 = (2x + 3)(2x – 3)
b
9 3
16x4 – 81
= (4x2 + 9) (2x + 3)(2x – 3)
2v2 - 72u2 = 2(v2 – 36u2) = 2[(v)2 – (6u)2] = 2(v + 6u)(v – 6u)
©2001 Michael Aryee
Factoring and Special Products
Page 1
Problem 1
a) x2 – 1
b) x2 – y2
c) 4x2 – y2
d) m4 –16
e) 9 – y6
f) 25x2 – 4a2
g) 100 – 49a6
h) s2 - 36
i) y2 – 49
j) 3a2 – 12x2
k) (2x + 3)2 – y2
l) (a+ b)2 – (a – b)2
m) x3 – x
n) 3ax2 – 3ay2
o) 100a2 – 64b2
p) s64 –t22
q) a4 – b2
r) x4 – 4y4
s) 9x2 – 64y2
t) 2x2 – 2y2
u) 44x4 – 99y4
v) 3a3 – 12a
w) a4 – b4
x) 16y12 – 1
©2001 Michael Aryee
Factoring and Special Products
Page 2

Factoring perfect squares
If the middle term of a trinomial is equal to twice the product of the positive square root of the
first term and the last terms, then that trinomial is a perfect square, and it is equal to the square of
the sum or difference of the positive square root of the first and last terms.
a2 + 2ab + b2 = (a + b)2
a2 - 2ab + b2 = (a - b)2
Note that middle term = twice the product of
a 2  a and
b2  b = 2ab.
Example
Factor each polynomial.
(a) x2 – 10x + 25
(b) 16x2 + 24x + 9
Solution:
(a)
x 2  x and
25 = 5 so that,
middle term = 2(x)(5) = 10x this match with the middle term of the original problem.
x2 – 10x + 25 = (x – 5)2
(b)
16x2  4 x and
9  3 so that,
middle term = 2(4x)(3) = 24x this match with the middle term of the original problem.
16x2 + 24x + 9 = (4x + 3)2
Problem
Factor each polynomial.
a) 9x2 – 30xy + 25y2
b) 4x2 + 36x + 81
c)
d) 16x2 – 40xy + 25y2
e) 16x2 + 56x + 49
f) x2 + 18x + 81
©2001 Michael Aryee
Factoring and Special Products
x2 + 24x + 144
Page 3
Sum and Difference of two cubes
The sum of two cubes is expressed as a3 + b3 = (first term)3 + (second term)3
Sum of two cubes
= [first term + second term][(first term)2 – (first term)(second term) + (second term)2]
In general,
a3 + b3 = (a + b)(a2 – ab + b2)
The difference of two cubes is expressed as a3 - b3 = (first term)3 - (second term)3
Difference of two cubes
= [first term - second term][(first term)2 + (first term)(second term) + (second term)2]
In general,
a3 - b3 = (a - b)(a2 + ab + b2)
Note that
3
a3  a and
3
b3  b .
If the original problem does not have an explicit base with an exponent of 3, use the cube root of
the expression to find the first and second terms. For example if 8x3 is given, the cube root of
8x3, ( 3 8x3 ) is 2x, so that we write (2x)3.
Example
Factor each polynomial.
(a) 8x3 – 27
(b) 64x6 + 1
Solution:
(a)
3
3
8x = 2x, and
3
27  3 so that,
a
2x
b
3
a2
(2x)2 = 4x2
ab
(2x)(3) = 6x
b2
(3)2 = 9
8x3 – 27 = (2x)3 – (3)3 = (2x – 3)[(2x)2 + (2x)(3) + (3)2]
= (2x – 3)(4x2 + 6x + 9)
(b)
3
64 x6  4 x2 , and 3 1  1 so that,
64x6 + 1
a
4x2
b
1
a2
(4x2)2 = 16x4
ab
2
(4x )(1) = 4x2
= (4x2)3 + (1)3 = (4x2 + 1)[(4x2)2 - (4x2)(1) + (1)2]
= (4x2 + 1)(16x4 - 4x2 + 1)
©2001 Michael Aryee
Factoring and Special Products
Page 4
b2
(1)2 = 1
Problem 2
a
b
a2
b2
ab
Factor each polynomial with the aid of this chart.
a) x3 – 1
b) x3 + y3
c) 27x3 – y3
d) m3 – 216
e) 27 – y3
f) 125x3 + 8a3
g) 1000 + 27a3
h) s3 - 64
i) y3 + 125
j) 3a3 – 81x2
k) (2x + 3)3 – y3
l) r3 + 8b3
m) x4 + x
n) 3ax3 – 3ay3
o) 54a3 – 128b3
p) s6 – t3
q) 128a3 – 2b3
r) x3 + 64y6
s) 81x3 + 64y3
t) 2x3 – 2y3
u) 81x3 – 3y3
©2001 Michael Aryee
Factoring and Special Products
Page 5