Factorization by Using Identities
Factorization by using identities will help us to factorize an algebraic expression easily.
The following identities are:
(i) (a + b)2 = a2 + 2ab +b2,(ii) (a - b)2 = a2 - 2ab + b2 and
(iii) a2 – b2 = (a + b)(a – b).
Now we will use these identities to factorize the given algebraic expressions.
Solved examples on factorization by using identities:
1. Factorize using the formula of square of the sum of two terms:
(i) z2 + 6z + 9
Solution:
= (z)2 + 2(z)(3) + (3)2
= (z + 3)2
= (z + 3)(z + 3)
(ii) x2 + 10x + 25
Solution:
We can express x2 + 10x + 25 as using a2 + 2ab + b2 = (a + b)2
= (x)2 + 2 ( x)( 5) + (5)2
= (x + 5)2
= (x + 5)(x - 5)
2. Factorize using the formula of square of the difference of two terms:
(i) 4m2 – 12mn + 9n2
Solution:
We can express 4m2 – 12mn + 9n2 as using a2 - 2ab + b2 = (a - b)2
= (2m)2 - 2(2m)(3n) + (3n)2
= (2m – 3n)2
= (2m - 3n)(2m - 3n)
(ii) x2 - 20x + 100
Solution:
We can express x2 - 20x + 100 as using a2 - 2ab + b2 = (a - b)2
= (x)2 - 2(x)(10) + (10)2
= (x - 10)2
=(x - 10)(x - 10)
3. Factorize using the formula of difference of two squares:
(i) 25x2 - 49
Solution:
We can express 25x2 - 49 as using a2 – b2 = (a + b)(a - b).
= (5x)2 - (7)2
= (5x + 7)(5x - 7)
(ii) 16x2 – 36y2
Solution:
We can express 16x2 – 36y2 as using a2 – b2 = (a + b)(a - b).
= (4x)2 - (6y)2
= (4x + 6y)(4x – 6y)
(iii) 1 – 25(2a – 5b)2
Solution:
We can express 1 – 25(2a – 5b)2 as using a2 – b2 = (a + b)(a - b).
= (1)2 - [5(2a – 5b)]2
= [1 + 5(2a – 5b)] [1 - 5(2a – 5b)]
= (1 + 10a – 25b) (1 – 10a + 25b)
4. Factor completely using the formula of difference of two squares: m4 – n4
Solution:
m4 – n4
We can express m4 – n4 as using a2 – b2 = (a + b)(a - b).
= (m2)2 - (n2)2
= (m2 + n2)( m2 - n2)
Now again, we can express m2 – n2 as using a2 – b2 = (a + b)(a - b).
= (m2 + n2) (m + n) (m - n)
8th Grade Math Practice
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