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Ex 14.3, 2 (iv) - Divide (x^3 + 2x^2 + 3x) Γ· 2x - Polynomial by Mono

Ex 14.3, 2 (iv) - Chapter 14 Class 8 Factorisation - Part 2
Ex 14.3, 2 (iv) - Chapter 14 Class 8 Factorisation - Part 3

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Ex 14.3, 2 (Method 1) Divide the given polynomial by the given monomial. (iv) (π‘₯^3 + 2π‘₯^2 + 3π‘₯) Γ· 2π‘₯ π‘₯^3 + 2π‘₯^2 + 3π‘₯ = (π‘₯^2 Γ— π‘₯) + (2π‘₯ Γ— π‘₯) + (3 Γ— π‘₯) Taking π‘₯ common, = π‘₯ (π‘₯^2 + 2π‘₯ + 3) Dividing (π‘₯^3 " + " 2π‘₯^2 " + " 3π‘₯)/2π‘₯ = (π‘₯ (π‘₯^2 + 2π‘₯ + 3))/2π‘₯ = π‘₯/π‘₯ Γ— (π‘₯^2+ 2π‘₯ + 3)/2 = (π‘₯^2+ 2π‘₯ + 3)/2 = 𝟏/𝟐 (𝒙^𝟐 + 2𝒙 + 3) Ex 14.3, 2 (Method 2) Divide the given polynomial by the given monomial. (iv) (π‘₯^3 + 2π‘₯^2 + 3π‘₯) Γ· 2π‘₯ (π‘₯^2+ 2π‘₯ + 3)/2π‘₯ = π‘₯^3/2π‘₯ + (2π‘₯^2)/2π‘₯ + 3π‘₯/2π‘₯ = (1/2Γ—π‘₯^3/π‘₯) + (2/2Γ—π‘₯^2/π‘₯) + (3/2Γ—π‘₯/π‘₯) = (1/2Γ—π‘₯^2 ) + (1Γ—π‘₯) + (3/2Γ—1) = 1/2 π‘₯^2+ π‘₯ + 3/2 = (π‘₯^2+ 2π‘₯ + 3 )/2 = 𝟏/𝟐 (𝒙^𝟐 + 2𝒙 + 3)

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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.