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Ex 14.3, 5 (vii) - Chapter 14 Class 8 Factorisation
Last updated at March 23, 2023 by Teachoo
Ex 14.3
Ex 14.3, 1 (i)
Ex 14.3, 1 (ii) Important
Ex 14.3, 1 (iii)
Ex 14.3, 1 (iv) Important
Ex 14.3, 1 (v)
Ex 14.3, 2 (i)
Ex 14.3, 2 (ii) Important
Ex 14.3, 2 (iii) Important
Ex 14.3, 2 (iv)
Ex 14.3, 2 (v) Important
Ex 14.3, 3 (i)
Ex 14.3, 3 (ii)
Ex 14.3, 3 (iii) Important
Ex 14.3, 3 (iv)
Ex 14.3, 3 (v) Important
Ex 14.3, 4 (i)
Ex 14.3, 4 (ii) Important
Ex 14.3, 4 (iii)
Ex 14.3, 4 (iv)
Ex 14.3, 4 (v) Important
Ex 14.3, 5 (i)
Ex 14.3, 5 (ii) Important
Ex 14.3, 5 (iii)
Ex 14.3, 5 (iv) Important
Ex 14.3, 5 (v)
Ex 14.3, 5 (vi)
Ex 14.3, 5 (vii) Important You are here
Transcript
Ex 14.3, 5 Factorise the expressions and divide them as directed. (vii) 39๐ฆ^3(50๐ฆ^2 โ 98) รท 26๐ฆ^2 (5y + 7) We first factorise ใ39๐ฆใ^3 (50๐ฆ^2โ 98) = ใ39๐ฆใ^3 (2รใ25๐ฆใ^2 โ 2 ร 49) Taking 2 common = ใ39๐ฆใ^3ร2 (ใ25๐ฆใ^2โ49) = ใ78๐ฆใ^3 (ใ25๐ฆใ^2โ49) = ใ78๐ฆใ^3 [(ใ5๐ฆใ^2)โ(7)^2] Using ๐^2 โ ๐^2 = (a + b) (a โ b) Here ๐= z and b = 4 Dividing 39๐ฆ^3(50๐ฆ^2 โ 98) รท 26๐ฆ^2 (5y + 7) = (39๐ฆ^3 (50๐ฆ^2 โ 98))/(26๐ฆ^2 (5๐ฆ + 7)) = (78๐ฆ^3 (5๐ฆ + 7) (5๐ฆ โ 7))/(26๐ฆ^2 (5๐ฆ + 7)) = 78/26 ร ๐ฆ^3/๐ฆ^2 ร ((5๐ฆ + 7))/((5๐ฆ + 7)) ร (5y โ 7) = 3 ร y ร (5y โ7) = 3y (5y โ 7)
