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

Factorization is a fundamental concept in algebra that helps break down complex expressions into simpler components. In Chapter 7 of RD Sharma for Class 8 students are introduced to the various methods of factorization including the grouping of terms using the identities and factorizing polynomials. Exercise 7.8 focuses on the application of these methods to solve different types of algebraic expressions.

Factorization

Factorization is the process of expressing a mathematical expression as a product of its factors. These factors are usually simpler or more manageable expressions. The primary goal is to break down complex algebraic expressions into the products of the binomials, trinomials or monomials which can be more easily manipulated. It involves techniques such as the taking out the greatest common factor, grouping and using the algebraic identities.

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)

Conclusion

The Factorization is a crucial skill in algebra, helping students simplify complex expressions and solve equations efficiently. Mastery of this concept lays the groundwork for the higher-level algebraic manipulations. Exercise 7.8 allows students to apply various techniques of the factorization and reinforces their understanding through the practice problems.