# RD Sharma Solutions for Class 8 Chapter - 7 Factorization Exercise 7.3

Students can refer to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization which are available here. The solutions here are solved step by step for a better understanding of the concepts, which helps students prepare for their exams at ease. Keeping in mind expert tutors at BYJUâ€™S have made this possible to help students crack difficult problems. Students can download the pdf of RD Sharma Solutions from the links provided below.

Exercise 7.3 of Chapter 7 Factorization is on based on the factorization of algebraic expressions when a binomial is a common factor.

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Factorize each of the following algebraic expressions:

1. 6x (2x – y) + 7y (2x – y)

Solution:

We have,

6x (2x – y) + 7y (2x – y)

By taking (2x â€“ y) as common we get,

(6x + 7y) (2x â€“ y)

2. 2r (y – x) + s (x – y)

Solution:

We have,

2r (y – x) + s (x – y)

By taking (-1) as common we get,

-2r (x â€“ y) + s (x â€“ y)

By taking (x – y) as common we get,

(x â€“ y) (-2r + s)

(x â€“ y) (s â€“ 2r)

3. 7a (2x – 3) + 3b (2x – 3)

Solution:

We have,

7a (2x – 3) + 3b (2x – 3)

By taking (2x – 3) as common we get,

(7a + 3b) (2x â€“ 3)

4. 9a (6a â€“ 5b) â€“ 12a2 (6a â€“ 5b)

Solution:

We have,

9a (6a â€“ 5b) â€“ 12a2 (6a â€“ 5b)

By taking (6a â€“ 5b) as common we get,

(9a â€“ 12a2) (6a â€“ 5b)

3a(3 â€“ 4a) (6a â€“ 5b)

5. 5 (x â€“ 2y)2 + 3 (x â€“ 2y)

Solution:

We have,

5 (x â€“ 2y)2 + 3 (x â€“ 2y)

By taking (x â€“ 2y) as common we get,

(x â€“ 2y) [5 (x â€“ 2y) + 3]

(x â€“ 2y) (5x â€“ 10y + 3)

6. 16 (2l â€“ 3m)2 – 12 (3m â€“ 2l)

Solution:

We have,

16 (2l â€“ 3m)2 – 12 (3m â€“ 2l)

By taking (-1) as common we get,

16 (2l â€“ 3m)2 + 12 (2l â€“ 3m)

By taking 4(2l â€“ 3m) as common we get,

4(2l â€“ 3m) [4 (2l â€“ 3m) + 3]

4(2l â€“ 3m) (8l â€“ 12m + 3)

7. 3a (x â€“ 2y) â€“ b (x â€“ 2y)

Solution:

We have,

3a (x â€“ 2y) â€“ b (x â€“ 2y)

By taking (x â€“ 2y) as common we get,

(3a â€“ b) (x â€“ 2y)

8. a2 (x + y) + b2 (x + y) + c2 (x + y)

Solution:

We have,

a2 (x + y) + b2 (x + y) + c2 (x + y)

By taking (x + y) as common we get,

(a2Â + b2Â + c2) (x + y)

9. (x – y)2 + (x – y)

Solution:

We have,

(x – y)2 + (x – y)

By taking (x – y) as common we get,

(x â€“ y) (x â€“ y + 1)

10. 6 (a + 2b) â€“ 4 (a + 2b)2

Solution:

We have,

6 (a + 2b) â€“ 4 (a + 2b)2

By taking (a + 2b) as common we get,

[6 â€“ 4 (a + 2b)] (a + 2b)

(6 â€“ 4a â€“ 8b) (a + 2b)

2(3 â€“ 2a â€“ 4b) (a + 2b)

11. a (x – y) + 2b (y – x) + c (x â€“ y)2

Solution:

We have,

a (x – y) + 2b (y – x) + c (x â€“ y)2

By taking (-1) as common we get,

a (x – y) – 2b (x – y) + c (x â€“ y)2

By taking (x – y) as common we get,

[a â€“ 2b + c(x – y)] (x – y)

(x – y) (a â€“ 2b + cx – cy)

12. -4 (x â€“ 2y)2 + 8 (x â€“ 2y)

Solution:

We have,

-4 (x â€“ 2y)2 + 8 (x â€“ 2y)

By taking 4(x – 2y) as common we get,

[-(x â€“ 2y) + 2] 4(x – 2y)

4(x – 2y) (-x + 2y + 2)

13. x3 (a â€“ 2b) + x2 (a â€“ 2b)

Solution:

We have,

x3 (a â€“ 2b) + x2 (a â€“ 2b)

By taking x2 (a â€“ 2b) as common we get,

(x + 1) [x2 (a â€“ 2b)]

x2 (a â€“ 2b) (x + 1)

14. (2x â€“ 3y) (a + b) + (3x â€“ 2y) (a + b)

Solution:

We have,

(2x â€“ 3y) (a + b) + (3x â€“ 2y) (a + b)

By taking (a + b) as common we get,

(a + b) [(2x â€“ 3y) + (3x â€“ 2y)]

(a + b) [2x -3y + 3x â€“ 2y]

(a + b) [5x â€“ 5y]

(a + b) 5(x – y)

15. 4(x + y) (3a – b) + 6(x + y) (2b â€“ 3a)

Solution:

We have,

4(x + y) (3a – b) + 6(x + y) (2b â€“ 3a)

By taking (x + y) as common we get,

(x + y) [4(3a – b) + 6(2b â€“ 3a)]

(x + y) [12a â€“ 4b + 12b â€“ 18a]

(x + y) [-6a + 8b]

(x + y) 2(-3a + 4b)

(x + y) 2(4b â€“ 3a)