Check sibling questions

Ex 14.2, 2 (viii)  - Factorise 25^a2 - 4b^2 + 28bc - 49^c2

Ex 14.2, 2 (viii) - Chapter 14 Class 8 Factorisation - Part 2


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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.