3 2 5ab 5b By taking 2 common in the first term and 5b as common in the second term 2 1 5 1 So we get a 1 2 5b By taking b as common a 1 b 2 5 So we get b a 1 2 5

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

Question 12 Factorization of Polynomials Exercise 3A

𝒂^𝟑𝒃 − 𝒂^𝟐𝒃 + 5ab – 5b

Answer:

𝑎^3𝑏 − 𝑎^2𝑏 + 5ab – 5b

By taking 𝑎^2𝑏 common in the first term and 5b as common in the second term

= 𝑎^2𝑏(𝑎 − 1) + 5𝑏(𝑎 − 1)

So we get,

= (a – 1) (𝑎^2𝑏 + 5b)

By taking b as common

= (a – 1) b (𝑎^2 + 5)

So we get,

= b (a – 1) (𝑎^2 + 5)

Related Questions

𝟗𝒙^𝟐 + 𝟏𝟐𝒙𝒚

𝟏𝟖𝒙^𝟐𝒚 − 𝟐𝟒 𝒙𝒚𝒛

𝟐𝟕𝒂^𝟑𝒃^𝟑 − 𝟒𝟓𝒂^𝟒𝒃^𝟐

𝟐𝒂 (𝒙 + 𝒚) − 𝟑𝒃(𝒙 + 𝒚)

2x (𝒑^𝟐 + 𝒒^𝟐) + 𝟒𝒚(𝒑^𝟐 + 𝒒^𝟐)

4(a+b) - 6(𝒂 + 𝒃)^𝟐

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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