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

Chapter 7 Factorization – Exercise 7.8 | Set 1

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)