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Ex 14.1, 2 (v) - Chapter 14 Class 8 Factorisation
Last updated at March 30, 2023 by Teachoo
Ex 14.1
Ex 14.1, 1 (i)
Ex 14.1, 1 (ii)
Ex 14.1, 1 (iii) Important
Ex 14.1, 1 (iv) Important
Ex 14.1, 1 (v)
Ex 14.1, 1 (vi) Important
Ex 14.1, 1 (vii)
Ex 14.1, 1 (viii) Important
Ex 14.1, 2 (i)
Ex 14.1, 2 (ii) Important
Ex 14.1, 2 (iii)
Ex 14.1, 2 (iv) Important
Ex 14.1, 2 (v) You are here
Ex 14.1, 2 (vi)
Ex 14.1, 2 (vii)
Ex 14.1, 2 (viii) Important
Ex 14.1, 2 (ix)
Ex 14.1, 2 (x) Important
Ex 14.1, 3 (i)
Ex 14.1, 3 (ii) Important
Ex 14.1, 3 (iii)
Ex 14.1, 3 (iv) Important
Ex 14.1, 3 (v) Important
Transcript
Ex 14.1, 2 (Method 1) Factorise the following expressions. (v) 20 𝑙^2 m + 30 a l m 20 𝑙^2 m = 20 × 𝑙^2 × m = 2 × 2 × 5 × 𝑙^2 × m = 2 × 2 × 5 × l × l × m 30 alm = 30 × a × l × m = 2 × 3 × 5 × a × l × m So, 20 𝑙^2m = 2 × 2 × 5 × l × l × m 30 alm = 2 × 3 × 5 × a × l × m So, 2, 5, l and m are the common factors. Now, 20 𝑙^2m + 30 alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m) Taking 2 × 5 × l × m common = 2 × 5 × l × m (2 × l + 3 × a) = 10lm(2l + 3a) Ex 14.1, 2 (Method 2) Factorise the following expressions. (v) 20 𝑙^2 m + 30 a l m 20 𝑙^2 m + 30 a l m = (10 × 2) 𝑙^2 m + (10 × 3) alm Taking 10 common, = 10 × (2𝑙^2m + 3 alm) = 10 × ((2l × lm) + (3a × lm)) Taking lm common, = 10 × lm (2l + 3a) = 10lm (2l + 3a)
