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Ex 14.2, 2 (viii) - Chapter 14 Class 8 Factorisation
Last updated at March 31, 2023 by Teachoo
Ex 14.2
Ex 14.2, 1 (i)
Ex 14.2, 1 (ii)
Ex 14.2, 1 (iii) Important
Ex 14.2, 1 (iv) Important
Ex 14.2, 1 (v)
Ex 14.2, 1 (vi)
Ex 14.2, 1 (vii) Important
Ex 14.2, 1 (viii)
Ex 14.2, 2 (i)
Ex 14.2, 2 (ii) Important
Ex 14.2, 2 (iii)
Ex 14.2, 2 (iv) Important
Ex 14.2, 2 (v)
Ex 14.2, 2 (vi)
Ex 14.2, 2 (vii) Important
Ex 14.2, 2 (viii) Important You are here
Ex 14.2, 3 (i)
Ex 14.2, 3 (ii) Important
Ex 14.2, 3 (iii)
Ex 14.2, 3 (iv) Important
Ex 14.2, 3 (v) Important
Ex 14.2, 3 (vi)
Ex 14.2, 3 (vii)
Ex 14.2, 3 (viii) Important
Ex 14.2, 3 (ix)
Ex 14.2, 4 (i)
Ex 14.2, 4 (ii) Important
Ex 14.2, 4 (iii)
Ex 14.2, 4 (iv) Important
Ex 14.2, 4 (v) Important
Ex 14.2, 5 (i)
Ex 14.2, 5 (ii) Important
Ex 14.2, 5 (iii)
Transcript
Ex 14.2, 2 Factorise. (viii) 25𝑎^2 – 4𝑏^2 + 28bc – 49𝑐^2 25𝑎^2 – 4𝑏^2 + 28bc – 49𝑐^2 Taking − common = 25𝑎^2 − (4𝑏^2 − 28bc +49𝑐^2) = 25𝑎^2 − (4𝑏^2 + 49𝑐^2 + 28bc) = 25𝑎^2 − ((2b)2 + (7c)2 − 2 × 2b × 7c) Using (x – y)2 = x2 + y2 – 2xy Here x = 2b and y = 7c = 25𝑎^2 − "(2b − " 〖"7c)" 〗^2 = (5a)2 − "(2b − " 〖"7c)" 〗^2 Using 𝑥^2 − 𝑦^2 = (𝑥 + y) (𝑥 − y) Here x = 5a and y = 2b − 7c Using 𝑥^2 − 𝑦^2 = (𝑥 + y) (𝑥 − y) Here x = 5a and y = 2b − 7c
