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Ex 12.2, 2 (iv) - 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 You are here
Ex 12.2, 2 (v)
Ex 12.2, 2 (vii) Important
Ex 12.2, 2 (viii) Important
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. (iv) 16π₯^5 β 144π₯^3 16π₯^5 β 144π₯^3 = γ16π₯γ^2 π₯^3β144π₯^3 Taking π₯^3 common, = π₯^3 (16π₯^2β144) = π₯^3 ((4π₯)^2β(12)^2 ) Using π^2β π^2 = (a + b) (a β b) Here a = 4x and b = 12 = π₯^3 (4π₯+12) (4π₯β12) Both have 4 as common factor = π₯^3 Γ 4 (π₯ + 3) Γ 4 (π₯ β 3) = π₯^3 Γ 4 Γ 4 (π₯ + 3) (π₯ β 3) = 16π^π (π + 3) (π β 3)
