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Ex 14.1, 2 (vi) - Chapter 14 Class 8 Factorisation
Last updated at March 23, 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)
Ex 14.1, 2 (vi) You are here
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. (vi) 5 𝑥^2 y – 15 〖𝑥𝑦〗^2 5 𝑥^2 y = 5 × 𝑥 × 𝑥 × y 15 𝑥y^2 = 15 × 𝑥 × 𝑦^2 = 3 × 5 × 𝑥 × 𝑦^2 = 3 × 5 × 𝑥 × 𝑦 × 𝑦 So, 5 𝑥^2 y = 5 × 𝑥 × 𝑥 × y 15𝑥^2 y = 3 × 5 × 𝑥 × 𝑥 × y So, 5, 𝑥 and y are the common factors. Now, 5 𝑥^2y – 15 〖𝑥𝑦〗^2 = (5 × 𝑥 × 𝑥 × y) − (3 × 5 × 𝑥 × y × y) Taking 5 × 𝑥 × y common = 5 × 𝑥 × y (𝑥 − (3 × y)) = 5xy (x − 3y) Ex 14.1, 2 (Method 2) Factorise the following expressions. (vi) 5 𝑥^2 y – 15 〖𝑥𝑦〗^2 5 𝑥^2y – 15 〖𝑥𝑦〗^2 = 5 𝑥^2y – 5 × 3 × 〖𝑥𝑦〗^2 Taking 5 × 𝑥 × y common = 5 (𝑥^2y − 3〖𝑥𝑦〗^2) = 5 ((𝑥𝑦 × 𝑥) − (𝑥𝑦 × 3y)) Taking xy common, = 5𝒙y (𝒙 − 3y)
