Difference of Squares – Explanation & Examples

The difference of two squares is a theorem that tells us if a quadratic equation can be written as a product of two binomials, in which one shows the difference of the square roots and the other shows the sum of the square roots.

One thing to note about this theorem is that it does not apply to the SUM of squares.

The difference of square formula is an algebraic form of the equation used to express the differences between two square values. A difference of square is expressed in the form:

a2 – b2, where both the first and last term is perfect squares. Factoring the difference of the two squares gives:

a2 – b2 = (a + b) (a – b)

This is true because, (a + b) (a – b) = a2 – ab + ab – b2 = a2 – b2

In this section, we will learn how to factorize algebraic expressions using the difference of square formula. To factor a difference of squares, the following steps are undertaken:

Example 1

Factor 64 – x2

Solution

Since we know the square of 8 is 64, then we can rewrite the expression as;64 – x2 = (8)2 – x2Now, apply the formula a2 – b2 = (a + b) (a – b) to factorize expression;= (8 + x) (8 – x).

Example 2

Factorize x 2 −16

Solution

Since x2−16 = (x) 2− (4)2, therefore apply the difference square formula a2 – b2 = (a + b) (a – b), where a and b in this case are x and 4 respectively.

Therefore, x2 – 42 = (x + 4) (x – 4)

Example 3

Factor 3a2 – 27b2

Solution

Since 3 is GCF of the terms, we factor it out.3a2 – 27b2 = 3(a2 – 9b2)=3[(a)2 – (3b)2]Now apply a2 – b2 = (a + b) (a – b) to get;= 3(a + 3b) (a – 3b)

Example 4

Factor x3 – 25xSolution

Since the GCF = x, factor it out;x3 – 25x = x (x2 – 25)= x (x2 – 52)Apply the formula a2 – b2 = (a + b) (a – b) to get;= x (x + 5) (x – 5).

Example 5

Factor the expression (x – 2)2 – (x – 3)2

Solution

In this problem a = (x – 2) and b = (x – 3)

We now apply a2 – b2 = (a + b) (a – b)

= [(x – 2) + (x – 3)] [(x – 2) – (x – 3)]

= [x – 2 + x – 3] [x – 2 – x + 3]

Combine the like terms and simplify the expressions;

[x – 2 + x – 3] [x – 2 – x + 3] = > [2x – 5] [1]

= [2x – 5]

Example 6

Factor the expression 25(x + y)2 – 36(x – 2y)2.

Solution

Rewrite the expression in the form a2 – b2.

25(x + y)2 – 36(x – 2y)2 => {5(x + y)}2 – {6(x – 2y)}2Apply the formula a2 – b2 = (a + b) (a – b) to get,

= [5(x + y) + 6(x – 2y)] [5(x + y) – 6(x – 2y)]

= [5x + 5y + 6x – 12y] [5x + 5y – 6x + 12y]

Collect like terms and simplify;

= (11x – 7y) (17y – x).

Example 7

Factor 2x2– 32.

Solution

Factor out the GCF;2x2– 32 => 2(x2– 16)= 2(x2 – 42)

Applying the difference squares formula, we get;= 2(x + 4) (x – 4)

Example 8

Factor 9x6 – y8

Solution

First, rewrite 9x6 – y8 in the form a2 – b2.

9x6 – y8 => (3x3)2 – (y4)2

Apply a2 – b2 = (a + b) (a – b) to get;

= (3x3 – y4) (3x3 + y4)

Example 9

Factor the expression 81a2 – (b – c)2

Solution

Rewrite 81a2 – (b – c)2 as a2 – b2= (9a)2 – (b – c)2By applying the formula of a2 – b2 = (a + b) (a – b) we get,= [9a + (b – c)] [9a – (b – c)]= [9a + b – c] [9a – b + c ]

Example 10

Factor 4x2– 25

Solution

= (2x)2– (5)2= (2x + 5) (2x – 5