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Factorisation using 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) You are here
Ex 12.1, 2 (viii) Important
Ex 12.1, 2 (ix)
Ex 12.1, 2 (x) Important
Last updated at May 29, 2023 by Teachoo
Ex 12.1, 2 (Method 1) Factorise the following expressions. (vii) 10 π^2 β 15 π^2 + 20 π^2 Now, 10π^2= 10 Γ π^2 = 2 Γ 5 Γ π^2 = 2 Γ 5 Γ a Γ a 15π^2 = 15 Γ π^2 = 3 Γ 5 Γ π^2 = 2 Γ 5 Γ b Γ b 20π^2 = 20 Γ π^2 = 2 Γ 2 Γ 5 Γ π^2 = 2 Γ 2 Γ 5 Γ c Γ c So, 5 is the common factor. 10π^2 β 15π^2 + 20π^2 = (2 Γ 5 Γ a Γ a) β (3 Γ 5 Γ b Γ b) + (2 Γ 2 Γ 5 Γ c Γ c) Taking 5 common = 5 Γ ((2 Γ a Γ a) β (3 Γ b Γ b) + (2 Γ 2 Γ c Γ c)) = 5 Γ (2π^2 β 3π^2 + 4π^2) = 5 (2π^π β 3π^π + 4π^π) Ex 12.1, 2 (Method 2) Factorise the following expressions. (vii) 10 π^2 β 15 π^2 + 20 π^2 10 π^2 β 15 π^2 + 20 π^2 = (5 Γ 2) π^2 β (5 Γ 3 ) π^2 + (5 Γ 4) π^2 Taking 5 common, = 5 (2π^π β 3π^π + 4π^π)