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Example 14 - Divide 24(x^2yz + xy^2z + xyz^2) by 8xyz using both

Example 14 - Chapter 14 Class 8 Factorisation - Part 2

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Example 14 (Method 1 – Separating each term) Divide 24(π‘₯^2yz + γ€–π‘₯𝑦〗^2z + γ€–π‘₯𝑦𝑧〗^2) by 8xyz using both the methods. 24 (π‘₯^2 𝑦𝑧+π‘₯𝑦^2 𝑧+π‘₯𝑦𝑧^2 ) = 24 (π‘₯Γ—π‘₯𝑦𝑧+𝑦×π‘₯𝑦𝑧+𝑧×π‘₯𝑦𝑧) Taking xyz common = 24 π‘₯𝑦𝑧 (π‘₯+𝑦+𝑧) Dividing (24 (π‘₯^2 𝑦𝑧 + π‘₯𝑦^2 𝑧 + π‘₯𝑦𝑧^2))/8π‘₯𝑦𝑧 = (24 π‘₯𝑦𝑧 (π‘₯ + 𝑦 + 𝑧))/8π‘₯𝑦𝑧 = (24 π‘₯𝑦𝑧 (π‘₯ + 𝑦 + 𝑧))/8π‘₯𝑦𝑧 = 3 (𝒙 + π’š + 𝒛) Example 14 (Method 2 – Canceling term) Divide 24(π‘₯^2yz + γ€–π‘₯𝑦〗^2z + γ€–π‘₯𝑦𝑧〗^2) by 8xyz using both the methods. (24 (π‘₯^2 𝑦𝑧 + π‘₯𝑦^2 𝑧 + π‘₯𝑦𝑧^2))/8π‘₯𝑦𝑧 = 24/8 Γ— ((π‘₯^2 𝑦𝑧 + π‘₯𝑦^2 𝑧 + π‘₯𝑦𝑧^2))/π‘₯𝑦𝑧 = 3 Γ— ((π‘₯^2 𝑦𝑧 + π‘₯𝑦^2 𝑧 + π‘₯𝑦𝑧^2))/π‘₯𝑦𝑧 = 3 Γ— ((π‘₯^2 𝑦𝑧 )/π‘₯𝑦𝑧+(π‘₯𝑦^2 𝑧)/π‘₯𝑦𝑧+(π‘₯𝑦𝑧^2)/π‘₯𝑦𝑧) = 3 Γ— (π‘₯ + 𝑦 + 𝑧) = 3 (𝒙 + π’š + 𝒛)

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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.