Accessing OP Malhotra Class 9 Solutions Chapter 4 Factorisation Ex 4(G) can be a valuable tool for students seeking extra practice.

## S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(G)

Factorise :

Question 1.
2x² + 3x + 1
Solution:
2x² + 3x + 1
= 2x² + 2x + x + 1

= 2x (x + 1) + 1 (x + 1)
= (x + 1) (2x + 1)

Question 2.
3x² + 17x + 10
Solution:
3x²+17x+10
= 3x² + 15x + 2x + 10
= 3x (x + 5) + 2 (x + 5)
= (x + 5) (3x + 2)

Question 3.
5x² + 9x – 2
Solution:
5x² + 9x – 2
= 5x² + 10x – x – 2

= 5x (x + 2) – 1 (x + 2)
= (x + 2) (5x – 1)

Question 4.
6x² – 7x – 5
Solution:
6x² – 7x – 5
= 6x² – 10x + 3x – 5

= 2x (3x – 5) + 1 (3x – 5)
= (3x – 5) (2x + 1)

Question 5.
6y² – 17y + 12
Solution:
6y² – 17y + 12
= 6y² – 9y – 8y + 12

= 3y – (2y – 3) – 4 (2y – 3)
= (2y – 3) (3y – 4)

Question 6.
8y² – 2y – 1
Solution:
8y² – 2y – 1
= 8y² – 4y + 2y – 1

= 4y (2y – 1) + 1 (2y – 1)
= (2y – 1) (4y + 1)

Question 7.
18bx² + 18bx – 20b
Solution:
18bx² + 18bx – 20b
= 2b (9x² + 9x – 10)

= 2b (9x² + 15x – 6x – 10)
= 2b [3x (3x + 5) – 2 (3x + 5)]
= 2b (3x + 5) (3x – 2)

Question 8.
14x² – 60xy + 16y²
Solution:
14x² – 60xy + 16y²
= 2 [7x² – 30xy + 8y²]
= 2 [7x² – 28xy – 2xy + 8y²]

= 2[7x(x – 4y) – 2y(x – 4y)]
= 2 (x – 4y) (7x – 2y)

Question 9.
30x² + 103xy – 7y²
Solution:
30x² + 103xy – 7y²
= 30x² + 105xy – 2xy – 7y²

= 15x (2x + 7y) – y (2x + 7y)
= (2x + 7y) (15x – y)

Question 10.
12x² – 29xy + 14y²
Solution:
12x² – 29xy + 14y²
= 12x² – 8xy – 21xy + 14y²

= 4x (3x – 2y) – 7y (3x – 2y)
= (3x – 2y) (4x – 7y)

Question 11.
15 + p (7 – 2p)
Solution:
15 + p (7 – 2p) = 15 + 7p – 2p²
= 15 + 10p – 3p – 2p²

= 5 (3 + 2p) – p (3 + 2p)
= (3 + 2p) (5 – p)

Question 12.
2 – y(7 – 5y)
Solution:
2 – y(7 – 5y)
= 2 – 7y + 5y²

= 2 – 5y – 2y + 5y²
= 1 (2 – 5y) – y(2 – 5y)
= (2 – 5y) (1 – y)

Question 13.
x(2x + 5) – 3
Solution:
x (2x + 5) – 3
= 2x² + 5x – 3

= 2x² + 6x – x – 3
= 2x (x + 3) – 1 (x + 3)
= (x + 3) (2x – 1)

Question 14.
1 – 18y – 63y²
Solution:
1 – 18y – 63y²
= 1 – 21y + 3y – 63y²

= 1 (1 – 21y) + 3y(1 – 21y)
= (1 – 21y) (1 + 3y)

Question 15.
8a³b – 10a²b² – 12ab³
Solution:
8a³b – 10a³b³ – 12ab³
= 2ab (4a² – 5ab – 6b²)
= 2ab [4a² – 8ab + 3ab – 6b²]

= 2ab [4a (a – 2b) + 3 b (a – 2b)]
= 2ab (a – 2b) (4a + 3b)

Question 16.
2 (a + b)² – 5 (a + b) – 3
Solution:
2 (a + b)² – 5 (a + b) – 3
Let a + b = x, then
2x² – 5x – 3 = 2x² – 6x + x – 3

= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)
= (a + b – 3) (2a + 2b + 1)
(Substituting the value of x)