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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8th Grade Math Practice

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