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![Factor completely: (i) 9x^2 + 24xy + 16y^2 [Class 9 Ganita Manjari I] - Exercise Set 4.2](https://cdn.teachoo.com/8f4ab6f0-a320-436f-941e-aeb057a71ac4/slide49.jpg)
Ex 4.2, 1
Last updated at May 15, 2026 by Teachoo
Class 9
Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)
- Consecutive Square Numbers
- Visualising Algebraic Identities
- Exercise Set 4.1
- Factorisation of Algebraic Expressions Using Identities
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Exercise Set 4.2
- More Identities
- Exercise Set 4.3
- Rewriting a2 - b2 Identity
- Factorisation Using Algebra Tiles
- Factorisation by Splitting the Middle term
- Exercise Set 4.4
- Finding New Algebraic Identities
- Simplifying Rational Expressions
- Exercise Set 4.5
- Word Problems
- End-of-Chapter Exercises
Transcript
Ex 4.2, 1 (i) Factor completely: (i) 9π₯^2+24π₯π¦+16π¦^2 Since 32 = 9 and 42 = 16 we can write our expression as 9π₯^2+24π₯π¦+16π¦^2 = 9π₯^2+16π¦^2+24π₯π¦ = (ππ)^π + (ππ)^π + 2 Γ ππ Γ ππ = (ππ+ππ)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = 3π₯, b = 4π¦ Ex 4.2, 1 (ii) Factor completely: (ii) 4π ^2+20π π‘+25π‘^2 Since 22 = 4 and 52 = 25 we can write our expression as 4π ^2+20π π‘+25π‘^2 = 4π ^2+25π‘^2+20π π‘ = (ππ)^π + (ππ)^π + 2 Γ ππ Γ ππ = (ππ+ππ)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = 2π , b = 5π‘ Ex 4.2, 1 (iii) Factor completely: (iii) 49π₯^2+28π₯π¦+4π¦^2 Since 72 = 49 and 22 = 4 we can write our expression as 49π₯^2+28π₯π¦+4π¦^2 = 49π₯^2+4π¦^2+28π₯π¦ = (ππ)^π + (ππ)^π + 2 Γ ππ Γ ππ = (ππ+ππ)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = 7π₯, b = 2π¦ Ex 4.2, 1 (iv) Factor completely: (iv) 64π^2+32/3 ππ+4/9 π^2 Since 82 = 64 and 22 = 4, 32 = 9 we can write our expression as 64π^2+32/3 ππ+4/9 π^2 = 64π^2+4/9 π^2+32/3 ππ = (ππ)^π + (π/π π)^π + 2 Γ ππ Γ π/π π = (ππ+π/π π)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = 8π, b = 2/3 π Ex 4.2, 1 (v) Factor completely: (v) 3π^2+4ππ+4/3 π^2 Here, 3 isnβt the square of any number And, 4/3 isnβt the square of any number So, to factorise this expression, we take some common factors out 3π^2+4ππ+4/3 π^2 = 3 Γ π^2+3 Γ4/3 ππ +3 Γ1/3 Γ4/3 π^2 = 3 Γ π^2+3 Γ4/3 ππ +3 Γ4/9 π^2 Taking 3 common = π Γ [π^π+π/π ππ+π/π π^π ] = 3 Γ [π^2+4/9 π^2+4/3 ππ] Since (π/π)^π=π/π and 12 = 1, we can factorise our expression = 3 Γ [π^π+(π/π π)^π+π Γ π Γπ/π π] = π Γ(π+π/π π)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = π, b = 2/3 π Ex 4.2, 1 (vi) Factor completely: (vi) 9/5 π ^2+6π π£+5π£^2 Here, π/π isnβt the square of any number, And, 5 isnβt the square of any number So, to factorise this expression, we take some common factors out 9/5 π ^2+6π π£+5π£^2 = 5 Γ1/5 Γ 9/5 π ^2+5 Γ1/5 Γ 6π π£+5 Γ π£^2 = 5 Γ 9/25 π ^2+5 Γ6/5 π π£+5 Γ π£^2 Taking 5 common = π Γ [π/ππ π^π+π/π ππ+π^π ] = 5 Γ [π/ππ π^π+π^π+π/π ππ] Since (π/π)^π=π/ππ and 12 = 1, we can factorise our expression = 5 Γ [(π/π π)^π+π^π+π Γπ/π π Γ π] = π Γ(π/π π+π)^π Using (π+π)^2 = π^2 + π^2 + 2ab Where π = π, b = 2/3 π
