Factor Binomials
The Difference of Two Squares
Factoring binomials is easy. If the binomial is the difference of two
perfect squares, it is factorable. If the binomial is the sum of two perfect
squares, it cannot be factored and is prime. (The sum and difference of
two cubes will be discussed later on.)
a2 – b2 = (a + b)(a – b)
a2 + b2 = Prime
Example 1: Factor: x2 – 16
We are subtracting, so it is a difference
Two perfect squares: x2 = x·x and 16 = 4·4
(x + 4)(x – 4)
To check the answer, multiply: (x + 4)(x – 4)
x(x – 4) + 4(x – 4)
x2 – 4x + 4x – 16
x2 + 0x – 16
x2 – 16 Correct
Note multiplication is commutative, so the order of the two binomial
factors doesn’t matter: (x – 4)(x + 4) will also equal x2 – 16.
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 2: Factor: x2 + 36
We are adding, so it is not the difference of two squares
Prime
Let’s try it, though:
x2 + 36 = x2 + 0x + 36
Factor of 36: 1,36; 2,18; 3,12; 4,9; 6,6
No set of factors add to 0, so the binomial is prime.
Example 3: Factor: x2 – 8
We are subtracting, so it is a difference
While x2 is a perfect square, 8 is not (4+4=8 sum not a product)
It is not the difference of two squares
Prime
Example 4: Factor: 9a2 – 25b2
We are subtracting, so it is a difference
Two perfect squares: 9a2 = 3a·3a and 25b2 = 5b·5b
(3a + 5b)(3a – 5b)
To check the answer, multiply: (3a + 5b)(3a – 5b)
3a(3a – 5b) + 5b(3a – 5b)
3a2 – 15ab + 15ab – 25b2
3a2 – 25b2 Correct
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
Example 5: Factor: y2 –
49
81
We are subtracting, so it is a difference
49
Two perfect squares: y2 = y·y and
7
81
7 7
= ·
9 9
7
(y + 9) (y − 9)
7
7
To check answer, multiply: (y + ) (y − )
9
9
7
7
7
y(y − ) + (y − )
9
9
9
y2 –
y2 –
7
7
49
9
81
y+ y–
9
49
81
Correct
Example 6: Factor: x4 – 1
We are subtracting, so it is a difference
Two perfect squares: x4 = x2·x2 and 1 = 1·1
(x2 + 1)(x2 – 1)
We still have a difference of two squares, (x2 – 1), factor further:
(x2 + 1)(x + 1)(x – 1)
To check the answer, multiply: (x2 + 1)(x + 1)(x – 1)
(x2 + 1)[(x + 1)(x – 1)]
(x2 + 1)[x(x – 1) + 1(x – 1)]
(x2 + 1)[x2 – 1x + 1x – 1]
(x2 + 1)(x2 – 1)
x2(x2 – 1) + 1(x2 – 1)
x4 – 1x2 + 1x2 – 1
x4 – 1
Correct
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)
When factoring, it is important not to forget about the GCF. If all terms
have a common factor, first factor out the GCF before factoring further.
Example 7: Factor: 3x3 – 12x
3x(x2 – 4)
Factor out the GCF, 3x, first
We are subtracting, so it is a difference
Two perfect squares: x2 = x·x and 4 = 2·2
3x(x + 2)(x – 2)
To check the answer, multiply: 3x(x + 2)(x – 2)
3x[(x + 2)(x – 2)]
3x[x(x – 2) + 2(x – 2)]
3x[x2 – 2x + 2x – 4]
3x[x2 – 4]
3x3 – 12x Correct
Example 8: Factor: x3 – 121
We are subtracting, so it is a difference
While 121 is a perfect square (121 = 11·11), x3 is not
There is no GCF, x, to factor out to get x2
It is not the difference of two squares
Prime
Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a
Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)