The availability of step-by-step OP Malhotra Class 9 Solutions Chapter 4 Factorisation Chapter Test can make challenging problems more manageable.

## S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Chapter Test

Factorise :

Question 1.

8x²y³ – x^{5}

Solution:

8x²y³ – x^{5} = x²(8y³ – x³)

= x²[(2y)³ – (x)³]

= x²(2y – x) (4y² + 2xy + x²)

Question 2.

x² + \(\frac { 1 }{ x² }\) + 2 – 2x – \(\frac { 2 }{ x }\)

Solution:

x² + \(\frac{1}{x^2}+2-2 x-\frac{2}{x}\)

= \(\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)\)

= \(\left(x+\frac{1}{x}\right)\left(x+\frac{1}{x}-2\right)\)

Question 3.

2x² – x – 6

Solution:

2x² – x – 6

⇒ 2x² – 4x + 3x – 6

= 2x(x – 2) + 3(x – 2)

= (x – 2) (2x + 3)

Question 4.

a³ – 0.216

Solution:

a³ – 0.216 = (a)³ – (0.6)³

= (a – 0.6) [a² + a x 0.6 + (0.6)²]

= (a – 0.6) (a² + 0.6a + 0.36)

Question 5.

6x²y – xy – 2y

Solution:

6x²y – xy – 2y

= y[6x² – x – 2]

= y[6x² – 4x + 3x – 2]

= y[2x(3x – 2) + 1(3x – 2)]

⇒ y(3x – 2) (2x + 1)

Question 6.

(x² – 3x)² – 8(x² – 3x) – 20

Solution:

(x² – 3x)² – 8(x² – 3x) – 20

Let x² – 3x = y,

then y² – 8y – 20

⇒ y(y – 10) + 2(y – 10)

⇒ (y – 10) (y + 2)

⇒ (x² – 3x – 10) (x² – 3x + 2)

⇒ {x² – 5x + 2x – 10} {x² – x – 2x + 2}

⇒ {x(x – 5) + 2(x – 5)} {x(x – 1) – 2(x – 1)}

⇒ (x – 5) (x + 2) (x – 1) (x – 2)

Multiple Choice Questions

Question 7.

One of the factors of (x – 1) – (x² – 1) is

(a) x² – 1

(b) x + 1

(c) x – 1

(d) x + 4

Solution:

(c) x – 1

(x – 1) – (x² – 1)

= (x – 1) – (x + 1) (x – 1)

= (x – 1) [1 – X + 1]

∴ x – 1 is its one factor.

Question 8.

If \(\frac{x}{y}+\frac{y}{x}\) = – 1 (x, y ≠ 0), then the value of x³ – y³ is

(a) 1

(b) – 1

(c) \(\frac { 1 }{ 2 }\)

(d) 0

Solution:

(d) 0

\(\frac{x}{y}+\frac{y}{x}\) = – 1

\(\frac{x^2+y^2}{x y}\) = – 1 ⇒ x² + y² = – xy

Now, x³ – y³ = (x – y) (x² + xy + y²)

= (x – y) x 0 = 0

Question 9.

The product of \(\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)\)

(a) x^{4} + \(\frac{1}{x^4}\)

(b) x³ + \(\frac { 1 }{ x³ }\) – 2

(c) x^{4} – \(\frac{1}{x^4}\)

(d) x² + \(\frac { 1 }{ x² }\) + 2

Solution:

(c) x^{4} – \(\frac{1}{x^4}\)

Question 10.

If x – 2y= 11 and xy = 8, then the value of x³ – 8y³ is

(a) 1860

(b) 1600

(c) 1859

(d) 2000

Solution:

x – 2y – 11, xy = 8

x – 2y = 11

Cubing both sides,

(x – 2y)³ = (11)³

⇒ x³ – 8y³ – 3 × x × (2y) (x – 2y) = 1331

⇒ x³ – 8y³ – 6xy(x – 2y) =1331

⇒ x³ – 8y³ – 6 x 8 x 11 = 1331

⇒ x³ – 8y³ – 528 = 1331

x³ – 8y³ = 1331 + 528 = 1859

∴ x³ – 8y³ = 1859