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Ex 14.3, 2 (v) - Divide (p^3q^6 - p^6q^3) Γ· p^3q^3 - Poly by Mono

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

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Ex 14.3, 2 (Method 1) Divide the given polynomial by the given monomial. (v) (𝑝^3 π‘ž^6 – 𝑝^6 π‘ž^3) Γ· 𝑝^3 π‘ž^3 𝑝^3 π‘ž^6 – 𝑝^6 π‘ž^3 = (𝑝^3 π‘ž^3 Γ— π‘ž^3) βˆ’ (𝑝^3 π‘ž^3 Γ— 𝑝^3) Taking 𝑝^3 π‘ž^3common, = 𝑝^3 π‘ž^3(π‘ž^3βˆ’π‘^3) Dividing (𝑝^3 π‘ž^6 " " βˆ’γ€– 𝑝〗^6 π‘ž^3)/(𝑝^3 π‘ž^3 ) = (𝑝^3 π‘ž^3 (π‘ž^3βˆ’π‘^3))/(𝑝^3 π‘ž^3 ) = 𝒒^πŸ‘βˆ’π’‘^πŸ‘ Ex 14.3, 2 (Method 2) Divide the given polynomial by the given monomial. (v) (𝑝^3 π‘ž^6 – 𝑝^6 π‘ž^3) Γ· 𝑝^3 π‘ž^3 (𝑝^3 π‘ž^6 " " βˆ’γ€– 𝑝〗^6 π‘ž^3)/(𝑝^3 π‘ž^3 ) = (𝑝^3 π‘ž^6 )/(𝑝^3 π‘ž^3 ) βˆ’ (𝑝^6 π‘ž^3)/(𝑝^3 π‘ž^3 ) = π‘ž^6/π‘ž^3 βˆ’π‘^6/𝑝^3 = π‘ž^(6 βˆ’ 3) βˆ’ 𝑝^(6 βˆ’ 3) = 𝒒^πŸ‘βˆ’π’‘^πŸ‘ ("As" π‘Ž^π‘š/π‘Ž^𝑛 =π‘Ž^(π‘š βˆ’ n) )

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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.