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• RD Sharma Class 8 Solutions for Maths

# Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.8 | Set 2

### Question 11. 12x2 – 17xy + 6y2

Solution:

Given:

12x2 – 17xy + 6y2

The coefficient of x2 = 12

The coefficient of x = -17y

Constant term = 6y2

So, we write the middle term -17xy as -9xy – 8xy

12x2 -17xy+ 6y2 = 12x2 – 9xy – 8xy + 6y2

= 3x (4x – 3y) – 2y (4x – 3y)

= (3x – 2y) (4x – 3y)

### Question 12. 6x2 – 5xy – 6y2

Solution:

Given:

6x2 – 5xy – 6y2

The coefficient of x2 = 6

The coefficient of x = -5y

Constant term = -6y2

So, we write the middle term -5xy as 4xy – 9xy

6x2 -5xy- 6y2 = 6x2 + 4xy – 9xy – 6y2

= 2x (3x + 2y) -3y (3x + 2y)

= (2x – 3y) (3x + 2y)

### Question 13. 6x2 – 13xy + 2y2

Solution:

Given:

6x2 – 13xy + 2y2

The coefficient of x2 = 6

The coefficient of x = -13y

Constant term = 2y2

So, we write the middle term -13xy as -12xy – xy

6x2 -13xy+ 2y2 = 6x2 – 12xy – xy + 2y2

= 6x (x – 2y) – y (x – 2y)

= (6x – y) (x – 2y)

### Question 14. 14x2 + 11xy – 15y2

Solution:

Given:

14x2 + 11xy – 15y2

The coefficient of x2 = 14

The coefficient of x = 11y

Constant term = -15y2

So, we write the middle term 11xy as 21xy – 10xy

14x2 + 11xy- 15y2 = 14x2 + 21xy – 10xy – 15y2

= 2x (7x – 5y) + 3y (7x – 5y)

= (2x + 3y) (7x – 5y)

### Question 15. 6a2 + 17ab – 3b2

Solution:

Given:

6a2 + 17ab – 3b2

The coefficient of a2 = 6

The coefficient of a = 17b

Constant term = -3b2

So, we write the middle term 17ab as 18ab – ab

6a2 +17ab– 3b2 = 6a2 + 18ab – ab – 3b2

= 6a (a + 3b) – b (a + 3b)

= (6a – b) (a + 3b)

### Question 16. 36a2 + 12abc – 15b2c2

Solution:

Given:

36a2 + 12abc – 15b2c2

The coefficient of a2 is 36

The coefficient of a is 12bc

Constant term is -15b2c2

So, we write the middle term 12abc as 30abc – 18abc

36a2 –12abc– 15b2c2 = 36a2 + 30abc – 18abc – 15b2c2

= 6a (6a + 5bc) – 3bc (6a + 5bc)

= (6a + 5bc) (6a – 3bc)

= (6a + 5bc) 3(2a – bc)

### Question 17. 15x2 – 16xyz – 15y2z2

Solution:

Given:

15x2 – 16xyz – 15y2z2

The coefficient of x2 = 15

The coefficient of x = -16yz

Constant term = -15y2z2

So, we write the middle term -16xyz as -25xyz + 9xyz

15x2 -16xyz- 15y2z2 = 15x2 – 25yz + 9yz – 15y2z2

= 5x (3x – 5yz) + 3yz (3x – 5yz)

= (5x + 3yz) (3x – 5yz)

### Question 18. (x – 2y)2 – 5 (x – 2y) + 6

Solution:

Given:

(x – 2y)2 – 5 (x – 2y) + 6

The coefficient of (x-2y)2 = 1

The coefficient of (x-2y) = -5

Constant term = 6

So, we write the middle term -5(x – 2y) as -2(x – 2y) -3(x – 2y)

(x – 2y)2 – 5 (x – 2y) + 6 = (x – 2y)2 – 2 (x – 2y) – 3 (x – 2y) + 6

= (x – 2y – 2) (x – 2y – 3)

### Question 19. (2a – b)2 + 2 (2a – b) – 8

Solution:

Given:

(2a – b)2 + 2 (2a – b) – 8

The coefficient of (2a-b)2 = 1

The coefficient of (2a-b) = 2

Constant term = -8

So, we write the middle term 2(2a – b) as 4 (2a –b) – 2 (2a – b)

(2a – b)2 + 2 (2a – b) – 8 = (2a – b)2 + 4 (2a – b) – 2 (2a – b) – 8

= (2a – b) (2a – b + 4) – 2 (2a – b + 4)

= (2a – b + 4) (2a – b – 2)