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Ex 14.3, 5 (vii) - Factorise and divide 39y^3 (50y^2 - 98) Γ· 26y2(5y+7

Ex 14.3, 5 (vii) - Chapter 14 Class 8 Factorisation - Part 2

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Ex 14.3, 5 Factorise the expressions and divide them as directed. (vii) 39𝑦^3(50𝑦^2 – 98) Γ· 26𝑦^2 (5y + 7) We first factorise γ€–39𝑦〗^3 (50𝑦^2βˆ’ 98) = γ€–39𝑦〗^3 (2Γ—γ€–25𝑦〗^2 βˆ’ 2 Γ— 49) Taking 2 common = γ€–39𝑦〗^3Γ—2 (γ€–25𝑦〗^2βˆ’49) = γ€–78𝑦〗^3 (γ€–25𝑦〗^2βˆ’49) = γ€–78𝑦〗^3 [(γ€–5𝑦〗^2)βˆ’(7)^2] Using π‘Ž^2 βˆ’ 𝑏^2 = (a + b) (a βˆ’ b) Here π‘Ž= z and b = 4 Dividing 39𝑦^3(50𝑦^2 – 98) Γ· 26𝑦^2 (5y + 7) = (39𝑦^3 (50𝑦^2 βˆ’ 98))/(26𝑦^2 (5𝑦 + 7)) = (78𝑦^3 (5𝑦 + 7) (5𝑦 βˆ’ 7))/(26𝑦^2 (5𝑦 + 7)) = 78/26 Γ— 𝑦^3/𝑦^2 Γ— ((5𝑦 + 7))/((5𝑦 + 7)) Γ— (5y βˆ’ 7) = 3 Γ— y Γ— (5y βˆ’7) = 3y (5y βˆ’ 7)

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