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Factorisation using common factors

Factorization by common factors

Ex 12.1, 1 (i)

Ex 12.1, 1 (ii)

Ex 12.1, 1 (iii) Important

Ex 12.1, 1 (iv) Important

Ex 12.1, 1 (v)

Ex 12.1, 1 (vi) Important

Ex 12.1, 1 (vii)

Ex 12.1, 1 (viii) Important

Example 2 Important

Example 1

Ex 12.1, 2 (i)

Ex 12.1, 2 (ii) Important

Ex 12.1, 2 (iii)

Ex 12.1, 2 (iv) Important

Ex 12.1, 2 (v)

Ex 12.1, 2 (vi)

Ex 12.1, 2 (vii)

Ex 12.1, 2 (viii) Important You are here

Ex 12.1, 2 (ix)

Ex 12.1, 2 (x) Important

Last updated at May 18, 2023 by Teachoo

Ex 12.1, 2 (Method 1) Factorise the following expressions. (viii) β 4π^2 + 4 ab β 4 ca Now, β 4π^2 = β4 Γ π^2 = β1 Γ 4 Γ π^2 = β1 Γ 2 Γ 2 Γ π^2 = β1 Γ 2 Γ 2 Γ a Γ a 4ab = 4 Γ a Γ b = 2 Γ 2 Γ a Γ b 4ca = 4 Γ c Γ a = 2 Γ 2 Γ c Γ a So, 2, 2 and a are the common factors. β 4π^2 + 4ab β4ca = (β1 Γ 2 Γ 2 Γ a Γ a) + (2 Γ 2 Γ a Γ b) β (2 Γ 2 Γ c Γ a) Taking 2 Γ 2 Γ a common, = 2 Γ 2 Γ a Γ ((β1 Γ a) + b β c) = 4a (βa + b β c) Ex 12.1, 2 (Method 2) Factorise the following expressions. (viii) β 4π^2 + 4 ab β 4 ca β 4π^2 + 4 ab β 4 ca Taking 4 common, = 4 (βπ^2 + ab β ca) = 4 ((βa Γ a) + (a Γ b) β (c Γ a)) Taking a common, = 4a (βa + b β c)