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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