62 32 7 2 3 20 Consider 2 3 a So we get 62 7 20 We can further write it as 62 15 8 20 By taking the common terms out 3a 2a 5 4 2a 5 3a 4 2a 5 Let us replace a by 2 3 So we get 32 3 4 22 3 5 6x 9 4 4x 6 5

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

Question 63 Factorization of Polynomials Exercise 3C

6(2𝑥 −3/𝑥)^2 + 7 (2𝑥 −3/𝑥) − 20

Answer:

6(2𝑥 −3/𝑥)^2 + 7 (2𝑥 −3/𝑥) − 20

Consider,

2𝑥 −3/𝑥= a

So we get,

= 6𝑎^2 + 7𝑎 − 20

We can further write it as

= 6𝑎^2 + 15𝑎 − 8𝑎 − 20

By taking the common terms out

= 3a (2a + 5) – 4 (2a + 5)

= (3a – 4) (2a + 5)

Let us replace a by 2𝑥 −3/𝑥

So we get,

(3(2𝑥 −3/𝑥) – 4) (2(2𝑥 −3/𝑥) + 5)

= (6x -9/𝑥– 4) (4x -6/𝑥+ 5)

Related Questions

𝑥^2 + 11𝑥 + 30

𝑥^2 + 18𝑥 + 32

𝑥^2 + 20𝑥 − 69

𝑥^2 + 19𝑥 − 150

𝑥^2 + 7𝑥 − 98

𝑥^2 + 2√3𝑥 − 24

Was This helpful?

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