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Example 2 - Chapter 12 Class 8 Factorisation
Last updated at May 18, 2023 by Teachoo
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 You are here
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
Ex 12.1, 2 (ix)
Ex 12.1, 2 (x) Important
Transcript
Example 2 (Method 1) Factorise 10π₯^2 β 18π₯^3 + 14π₯^4 10π₯^2 = 10 Γ π₯^2 = 2 Γ 5 Γ π₯^2 = 2 Γ 5 Γ π₯ Γ π₯ 18π₯^3 = 18 Γ π₯^3 = 2 Γ 3 Γ 3 Γ π₯^3 = 2 Γ 3 Γ 3 Γ π₯ Γ π₯ Γ π₯ 14π₯^4 = 14 Γ π₯^4 = 2 Γ 7 Γ π₯^4 = 2 Γ 7 Γ π₯ Γ π₯ Γ π₯ Γ π₯ Now, 10π₯^2 = 2 Γ 5 Γ π₯ Γ π₯ 18π₯^3 = 2 Γ 3 Γ 3 Γ π₯ Γ π₯ Γ π₯ 14π₯^2 = 2 Γ 7 Γ π₯ Γ π₯ Γ π₯ Γ π₯ So, 2, π₯, π₯ are common factors. 10π₯^2 + 18π₯^3+14π₯^4 = (2 Γ 5 Γ π₯ Γ π₯) β (2 Γ 3 Γ 3 Γ π₯ Γ π₯" Γ " π₯) + (2 Γ 7 Γ π₯ Γ π₯ Γ π₯ Γ π₯) = 2 Γ π₯ Γ π₯ (5 β (3 Γ 3 Γ π₯) + (7 Γ π₯ Γ π₯)) = 2π₯^2 (5 β 9π₯ + 7π₯^2) = 2π^π (7π^π β 9π + 5) Example 2 (Method 2) Factorise 10π₯^2 β 18π₯^3 + 14π₯^410π₯^2 β 18π₯^3 + 14π₯^4 = (2 Γ 5π₯^2) β (2 Γ 9π₯^3) + (2 Γ 7π₯^4) Taking 2 common from all the terms = 2 (5π₯^2 β 9π₯^3 + 7π₯^4) = 2 ( (5 Γ π₯^2) β (9π₯ Γ π₯^2) + (7π₯^2 Γ π₯^2) ) Taking π₯^2 common from all the terms = 2π₯^2 (5 β 9π₯ + 7π₯^2) = 2π^π (7π^π β 9π + 5)
