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Factorization notes of class 8 CBSE by Abhinav Sir

by Abhinav Anand on June 20, 2025 in 8M, Maths, Notes8, Online

Factorisation Notes for Class 8 CBSE with Solutions

Introduction to Factorisation

Factorisation is the process of finding two or more expressions whose product is the given expression. It is the reverse process of multiplication.

Factors: When an algebraic expression can be written as the product of two or more expressions, then each of these expressions is called a factor of the given expression.
Factorisation: The process of finding two or more expressions whose product is the given expression is called factorisation.

1. Factorisation When a Common Monomial Factor Occurs in Each Term

Method:

Find the HCF of all the terms of the given expression.
Divide each term of the given expression by this HCF.
Write the given expression as the product of this HCF and the quotient obtained in step 2.

2. Factorisation by Grouping

Sometimes the terms of the given expression need to be arranged in suitable groups such that all the groups have a common factor.

Method:

Arrange the terms of the given expression in groups in such a way that all the groups have a common factor.
Factorise each group.
Take out the factor which is common to all such groups.

3. Factorisation When Given Expression is Difference of Two Squares

Formula: \( a^2 - b^2 = (a + b)(a - b) \)

Exercise 7A

Factorise:

(i) \( 12x + 15 \)
(ii) \( 16a^2 - 24ab \)
(iii) \( 9x^2 - 36x^2y \)
(iv) \( 14x^2 + 21x^2y - 28x^2y^2 \)
(v) \( 6x + 3 - 5lx + 5l \)
(vi) \( 6a(a - 2b) + 5b(a - 2b) \)
(vii) \( (x + 5)^2 - 4(x + 5) \)
(viii) \( 16(2p - 3q)^2 - 4(2p - 3q) \)
(ix) \( ab^2 - bc^2 + ab - bc \)
(x) \( a^2 + bc - ab - ac \)
(xi) \( 15m - 21 \)
(xii) \( 10x^2 - 15x \)
(xiii) \( 8x^2 - 72xy + 12x \)
(xiv) \( 25 - 10t + 20t^2 \)
(xv) \( 5x(x - 4) - 7x - 4 \)
(xvi) \( x^2(2a - b) - x^2(2a - b) \)
(xvii) \( 9a(3a - 5b) - 12a^2(3a - 5b) \)
(xviii) \( 2m(1 - n) + 3(1 - n) \)
(xix) \( 9a(3a - 5b) - 12a^2(3a - 5b) \)
(xx) \( 12(2x - 3y)^2 - 16(3y - 2x) \)

Solutions:

(i) \( 12x + 15 = 3(4x + 5) \)
(ii) \( 16a^2 - 24ab = 8a(2a - 3b) \)
(iii) \( 9x^2 - 36x^2y = 9x^2(1 - 4y) \)
(iv) \( 14x^2 + 21x^2y - 28x^2y^2 = 7x^2(2 + 3y - 4y^2) \)
(v) \( 6x + 3 - 5lx + 5l = 3(2x + 1) + 5l(1 - x) \)
(vi) \( 6a(a - 2b) + 5b(a - 2b) = (a - 2b)(6a + 5b) \)
(vii) \( (x + 5)^2 - 4(x + 5) = (x + 5)(x + 1) \)
(viii) \( 16(2p - 3q)^2 - 4(2p - 3q) = 4(2p - 3q)[4(2p - 3q) - 1] \)
(ix) \( ab^2 - bc^2 + ab - bc = (b + 1)(ab - bc) \)
(x) \( a^2 + bc - ab - ac = (a - b)(a - c) \)
(xi) \( 15m - 21 = 3(5m - 7) \)
(xii) \( 10x^2 - 15x = 5x(2x - 3) \)
(xiii) \( 8x^2 - 72xy + 12x = 4x(2x - 18y + 3) \)
(xiv) \( 25 - 10t + 20t^2 = 5(4t^2 - 2t + 5) \)
(xv) \( 5x(x - 4) - 7x - 4 = 5x^2 - 27x - 4 \) (Further factoring needed if quadratic method is allowed)
(xvi) \( x^2(2a - b) - x^2(2a - b) = 0 \)
(xvii) \( 9a(3a - 5b) - 12a^2(3a - 5b) = 3a(3a - 5b)(3 - 4a) \)
(xviii) \( 2m(1 - n) + 3(1 - n) = (1 - n)(2m + 3) \)
(xix) \( 9a(3a - 5b) - 12a^2(3a - 5b) = 3a(3a - 5b)(3 - 4a) \)
(xx) \( 12(2x - 3y)^2 - 16(3y - 2x) = 4(2x - 3y)(6x - 9y + 4) \)

About Abhinav Anand
I am an exceptional teacher, passionate and approachable. I simplify complex ideas with clarity and creativity, fostering a supportive classroom where students thrive. My patience and enthusiasm inspire confidence, and I encourage curiosity while making every student feel valued.