# Factoring Differences of Squares

How to solve factoring differences of squares?

To factorize an algebraic expression expressible as the difference of two squares, we use the following identity a2 - b2 = (a + b) (a – b).

Solved examples on factoring differences of squares:

1. Factorize the following algebraic expressions:

(i) 64 - x2

Solution:

64 - x2

= (8)2 - x2, since we know 64 = 8 times 8 which is 82

Now by using the formula of a2 - b2 = (a + b)(a – b) to complete the factor fully.

= (8 + x)(8 - x).

(ii) 3a2 - 27b2

Solution:

3a2 - 27b2

= 3(a2 – 9b2), here we took 3 as common.

=3[(a)2 – (3b)2], since we know 9b2 = 3b times 3b which is (3b)2

So, now we need to apply the formula of a2 - b2 = (a + b)(a – b) to complete the factor fully.

= 3(a + 3b)(a – 3b)

(iii) x3 - 25x

Solution:

x3 - 25x

= x(x2 - 25), here we took x as common.

= x(x2 - 52), since we know, 25 = 52

Now we can write x2 – 52 as using the formula of a2 - b2 = (a + b)(a – b).

= x(x + 5)(x - 5).

2. Factor the expressions:

(i) 81a2 - (b - c)2

Solution:

We can write 81a2 - (b - c)2 as a2 - b2

= (9a)2 - (b - c)2, since we know, 81a2 = (9a)2

Now using the formula of a2 – b2 = (a + b) (a – b) we get,

= [9a + (b – c)] [9a - (b – c)]

= [9a + b – c] [9a - b + c ]

(ii) 25(x + y)2 - 36(x - 2y)2.

Solution:

We can write 25(x + y)2 - 36(x - 2y)2 as a2 - b2.

= {5(x + y)}2 - {6(x - 2y)}2

Now using the formula of a2 – b2 = (a + b) (a – b) we get,

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

= [5x + 5y + 6x – 12y] [5x + 5y – 6x + 12y], (applying distributive property)

Now we will arrange and then simplify it.

= (11x - 7y) (17y - x).

(iii) (x – 2)2 – (x – 3)2

Solution:

We can express (x – 2)2 – (x – 3)2 using the formula of a2 - b2 = (a + b) (a – b)

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

= [x – 2 + x - 3][x - 2 – x + 3]

Now we will arrange and then simplify it.

= [2x – 5][1]

= [2x – 5]

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