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)