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:

We can express z2 + 6z + 9 as using a2 + 2ab + b2 = (a + b)2

= (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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