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

### RD Sharma Class 8 Solutions Chapter 7 Ex 7.3 PDF Free Download

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 BYJU’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.

## Download the pdf of RD Sharma For Class 8 Maths Exercise 7.3 Chapter 7 Factorization    ### Access other Exercises of RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization

Exercise 7.1 Solutions

Exercise 7.2 Solutions

Exercise 7.4 Solutions

Exercise 7.5 Solutions

Exercise 7.6 Solutions

Exercise 7.7 Solutions

Exercise 7.8 Solutions

Exercise 7.9 Solutions

### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.3 Chapter 7 Factorization

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 – y) as common we get,

(6x + 7y) (2x – 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 – y) + s (x – y)

By taking (x – y) as common we get,

(x – y) (-2r + s)

(x – y) (s – 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 – 3)

4. 9a (6a – 5b) – 12a2 (6a – 5b)

Solution:

We have,

9a (6a – 5b) – 12a2 (6a – 5b)

By taking (6a – 5b) as common we get,

(9a – 12a2) (6a – 5b)

3a(3 – 4a) (6a – 5b)

5. 5 (x – 2y)2 + 3 (x – 2y)

Solution:

We have,

5 (x – 2y)2 + 3 (x – 2y)

By taking (x – 2y) as common we get,

(x – 2y) [5 (x – 2y) + 3]

(x – 2y) (5x – 10y + 3)

6. 16 (2l – 3m)2 – 12 (3m – 2l)

Solution:

We have,

16 (2l – 3m)2 – 12 (3m – 2l)

By taking (-1) as common we get,

16 (2l – 3m)2 + 12 (2l – 3m)

By taking 4(2l – 3m) as common we get,

4(2l – 3m) [4 (2l – 3m) + 3]

4(2l – 3m) (8l – 12m + 3)

7. 3a (x – 2y) – b (x – 2y)

Solution:

We have,

3a (x – 2y) – b (x – 2y)

By taking (x – 2y) as common we get,

(3a – b) (x – 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,

(a2 + b2 + 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 – y) (x – y + 1)

10. 6 (a + 2b) – 4 (a + 2b)2

Solution:

We have,

6 (a + 2b) – 4 (a + 2b)2

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

[6 – 4 (a + 2b)] (a + 2b)

(6 – 4a – 8b) (a + 2b)

2(3 – 2a – 4b) (a + 2b)

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

Solution:

We have,

a (x – y) + 2b (y – x) + c (x – y)2

By taking (-1) as common we get,

a (x – y) – 2b (x – y) + c (x – y)2

By taking (x – y) as common we get,

[a – 2b + c(x – y)] (x – y)

(x – y) (a – 2b + cx – cy)

12. -4 (x – 2y)2 + 8 (x – 2y)

Solution:

We have,

-4 (x – 2y)2 + 8 (x – 2y)

By taking 4(x – 2y) as common we get,

[-(x – 2y) + 2] 4(x – 2y)

4(x – 2y) (-x + 2y + 2)

13. x3 (a – 2b) + x2 (a – 2b)

Solution:

We have,

x3 (a – 2b) + x2 (a – 2b)

By taking x2 (a – 2b) as common we get,

(x + 1) [x2 (a – 2b)]

x2 (a – 2b) (x + 1)

14. (2x – 3y) (a + b) + (3x – 2y) (a + b)

Solution:

We have,

(2x – 3y) (a + b) + (3x – 2y) (a + b)

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

(a + b) [(2x – 3y) + (3x – 2y)]

(a + b) [2x -3y + 3x – 2y]

(a + b) [5x – 5y]

(a + b) 5(x – y)

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

Solution:

We have,

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

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

(x + y) [4(3a – b) + 6(2b – 3a)]

(x + y) [12a – 4b + 12b – 18a]

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

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

(x + y) 2(4b – 3a)

#### 1 Comment

1. Soyel rana

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