23 163 5 10 23 83 5 2 According to the equation 3 3 2 2 So we get 23 83 5 2 We can write it as 23 23 5 2 2 22 2 22 5 2 2 22 2 42 5 2 By taking 2as common We get 222 2 42 5

RS Aggarwal Solutions Class 9 Mathematics Solutions for Factorization of Polynomials Exercise 3F in Chapter 3 - Factorization of Polynomials

Question 27 Factorization of Polynomials Exercise 3F

2𝑎^3 + 16𝑏^3 − 5𝑎 − 10𝑏

Answer:

2𝑎^3 + 16𝑏^3 − 5𝑎 − 10𝑏

= 2(𝑎^3 + 8𝑏^3) − 5(𝑎 + 2𝑏)

According to the equation,

𝑎^3 + 𝑏^3 = (𝑎 + 𝑏)(𝑎^2 − 𝑎𝑏 + 𝑏^2)

So we get,

2(𝑎^3 + 8𝑏^3) − 5(𝑎 + 2𝑏)

We can write it as,

= 2(𝑎^3 + (2𝑏)^3) − 5(𝑎 + 2𝑏)

= 2(𝑎 + 2𝑏)(𝑎^2 − 𝑎(2𝑏) + (2𝑏)^2) −5(𝑎 + 2𝑏)

= 2(𝑎 + 2𝑏)(𝑎^2 − 2𝑎𝑏 + 4𝑏^2) − 5(𝑎 + 2𝑏)

By taking (𝑎 + 2𝑏)as common,

We get,

= (𝑎 + 2𝑏)(2(𝑎^2 − 2𝑎𝑏 + 4𝑏^2) − 5)

Related Questions

27𝑎^3 + 64𝑏^3

8𝑥^3 −1/27𝑦^3

𝑥^3/216− 8𝑦^3

3𝑎^7𝑏 − 81𝑎^4𝑏^4

𝑥^4𝑦^4 − 𝑥𝑦

8𝑥^2𝑦^3 − 𝑥^5

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Exercises

Factorization of Polynomials Exercise 3A

Factorization of Polynomials Exercise 3B

Factorization of Polynomials Exercise 3C

Factorization of Polynomials Exercise 3D

Factorization of Polynomials Exercise 3E

Factorization of Polynomials Exercise 3F

Factorization of Polynomials Exercise 3G

Chapters

Number Systems

Polynomials

Factorization of Polynomials

Linear Equations In Two Variables

Coordinate Geometry

Introduction To Euclids Geometry

Lines And Angles

Triangles

Congruence Of Triangles And Inequalities In A Triangle

Quadrilaterals

Areas Of Parallelograms And Triangles

Circles

Geometrical Constructions

Areas Of Triangles And Quadrilaterals

Volume And Surface Area Of Solids

Mean Median And Mode Of Ungrouped Data

Probability