Ch- Test of Factorisation Class 9 OP Malhotra ICSE Maths Solutions Ch-4.  We Provide Step by Step Solutions / Answer of Factorisation OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Factorisation Class 9 OP Malhotra Ch-Test ICSE Maths Solutions Ch-4

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-4	Factorisation
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Factorisation

Factorisation Class 9 OP Malhotra ICSE Maths Solutions Ch-4

Que-1: 8x²y³ – x^5

Sol: 8x²y³ – x^5 = x²(8y³ – x³)
= x²[(2y)³ – (x)³]
= x²(2y – x) (4y² + 2xy + x²)

Que-2: x² + (1/x²) + 2 – 2x – (2/x)

Sol: x² + (1/x²) + 2 − 2x − (2/x)
= (x+1/x)² − 2(x+1/x)
= (x + 1/x) (x + 1/x − 2)

Que-3: 2x² – x – 6

Sol: 2x² – x – 6
⇒ 2x² – 4x + 3x – 6
= 2x(x – 2) + 3(x – 2)
= (x – 2) (2x + 3)

Que-4: a³ – 0.216

Sol: 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)

Que-5: 6x²y – xy – 2y

Sol: 6x²y – xy – 2y
= y[6x² – x – 2]
= y[6x² – 4x + 3x – 2]
= y[2x(3x – 2) + 1(3x – 2)]
⇒ y(3x – 2) (2x + 1)

Que-6:(x² – 3x)² – 8(x² – 3x) – 20

Sol: (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

Que-7: One of the factors of (x – 1) – (x² – 1) is

(a) x² – 1
(b) x + 1
(c) x – 1
(d) x + 4

Sol: (c) x – 1
(x – 1) – (x² – 1)
= (x – 1) – (x + 1) (x – 1)
= (x – 1) [1 – X + 1]
∴ x – 1 is its one factor.

Que-8: If (x/y)+(y/x) = – 1 (x, y ≠ 0), then the value of x³ – y³ is

(a) 1
(b) – 1
(c) 12
(d) 0

Solution: (d) 0
(x/y)+(y/x) = – 1
(x²+y²)/xy = – 1
⇒ x² + y² = – xy
Now, x³ – y³ = (x – y) (x² + xy + y²)
= (x – y) x 0 = 0

Que-9: The product of (x − 1/x) (x + 1/x) (x² + 1/x²)

(a) x^4 + (1/x^4)
(b) x³ + (1/x³) – 2
(c) x^4 – (1/x^4)
(d) x² + (1/x²) + 2

Solution: (c) x^4 – (1/x^4)
(x − 1/x) (x + 1/x) (x² + 1/x²)
(x² – 1/x²) (x² + 1/x²)
(x²)² – {(1/x²)}²
= x^4 – (1/x^4)

Que-10: If x – 2y= 11 and xy = 8, then the value of x³ – 8y³ is

(a) 1860
(b) 1600
(c) 1859
(d) 2000

Sol: 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

— : End of Factorisation 9 OP Malhotra Ch-Test ICSE Maths Step by step Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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