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Ex 12.2, 2 (viii) - Chapter 12 Class 8 Factorisation
Last updated at May 29, 2023 by Teachoo
Factorisation using identities
Factorisation using identities
Example 4
Example 5 Important
Example 6
Example 8
Example 7 Important
Ex 12.2, 1 (i)
Ex 12.2, 1 (ii)
Ex 12.2, 1 (iii) Important
Ex 12.2, 1 (v)
Ex 12.2, 1 (viii)
Ex 12.2, 1 (vi)
Ex 12.2, 1 (iv) Important
Ex 12.2, 1 (vii) Important
Ex 12.2, 2 (i)
Ex 12.2, 2 (iii)
Ex 12.2, 2 (vi)
Ex 12.2, 2 (ii) Important
Ex 12.2, 2 (iv) Important
Ex 12.2, 2 (v)
Ex 12.2, 2 (vii) Important
Ex 12.2, 2 (viii) Important You are here
Ex 12.2, 4 (i)
Ex 12.2, 4 (ii) Important
Ex 12.2, 4 (iii)
Ex 12.2, 4 (iv) Important
Ex 12.2, 4 (v) Important
Transcript
Ex 12.2, 2 Factorise. (viii) 25π^2 β 4π^2 + 28bc β 49π^2 25π^2 β 4π^2 + 28bc β 49π^2 Taking β common = 25π^2 β (4π^2 β 28bc +49π^2) = 25π^2 β (4π^2 + 49π^2 + 28bc) = 25π^2 β ((2b)2 + (7c)2 β 2 Γ 2b Γ 7c) Using (x β y)2 = x2 + y2 β 2xy Here x = 2b and y = 7c = 25π^2 β "(2b β " γ"7c)" γ^2 = (5a)2 β "(2b β " γ"7c)" γ^2 Using π₯^2 β π¦^2 = (π₯ + y) (π₯ β y) Here x = 5a and y = 2b β 7c Using π₯^2 β π¦^2 = (π₯ + y) (π₯ β y) Here x = 5a and y = 2b β 7c
