# RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3

In this chapter, we provide RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3 Maths pdf, free RD Sharma Class 8 Solutions Chapter 7 Factorization Ex 7.3 Maths book pdf download. Now you will get step by step solution to each question.

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

Factorize each of the following algebraic expressions.

Question 1.
6x (2x – y) + 7y (2x – y)
Solution:
6x (2x – y) + 7y (2x – y)
= (2x – y) (6x + 7y)
[∵ (2x – y) is common]

Question 2.
2r (y – x) + s (x – y)
Solution:
2r (y – x) + s (x – y)
-2r (x – y) +s (x – y)
= (x – y) (-2r + s)                   [(x – y) is common]
= (x-y) (s-2r)

Question 3.
7a (2x – 3) + 2b (2x – 3)

Solution:
7a (2x – 3) + 3b (2x – 3)
= (2x – 3) (7a + 3b)               [(2x – 3) is common]

Question 4.
9a (6a – 5b) – 12a2 (6a – 5b)
Solution:
9a (6a – 5b) – 12a2 (6a – 5b)
HCF of 9 and 12 = 3
∴ 3a (6a – 5b) (3 – 4a)
{(6a – 5b) is common}

Question 5.
5 (x – 2y)2 + 3 (x – 2y)

Solution:
5 (x – 2y)2 + 3 (x – 2y)
= 5 (x – 2y) (x – 2y) + 3 (x – 2y)
= (x – 2y) {5 (x – 2y) + 3}
{(x – 2y) is common}
= (x – 2y) (5x – 10y + 3)

Question 6.
16 (2l – 3m)2 – 12 (3m – 2l)
Solution:
16 (2l – 3m)2 – 12 (3m-2l)
= 16 (2l – 3m) (2l – 3m) + 12 (2l – 3m)
HCF of 16, 12 = 4 4 (2l-3m) {4 (2l- 3m) + 3}
{(2l – 3m) is common}
= 4 (2l -3m) (8l- 12m+ 3)

Question 7.
3a (x – 2y) – b (x – 2y)
Solution:
3a (x – 2y) – b (x – 2y)
= (x – 2y) (3a – b)
{(x – 2y) is common}

Question 8.
a2 (x + y) + b2 (x + y) + c2 (x + y)
Solution:
a2 (x + y) + b2 (x + y) + c2 (x + 3’)
= (x + y) (a2 + b2 + c2)
{(x + y) is common}

Question 9.
(x-y)2 + (x -y)

Solution:
(x – y)2 + (x- y) = (x – y) (x – y) + (x – y)
= (x – y) (x – y + 1)                          {(a – y) is common}

Question 10.
6 (a + 2b) – 4 (a + 2b)2
Solution:
6 (a + 2b) – 4 (a + 2b)2
= 6 (a + 2b) – 4 (a + 2b) (a + 2b)
HCF of 6, 4 = 2
= 2 {a + 2b) {3 – 2 {a + 2b)
{2 (a + b) is common}
= 2 (a + 2b) (3-2 a- 4b)

Question 11.
a (x -y) + 2b (y – x) + c (x -y)2
Solution:
a (x -y) + 2b (y – x) + c (x -y)2
= a (x – y) – 2b (x – y) + c (x – y) {x – y)
= (x – y) {x – 2b + c (x – y)}
{(a – y) is common}
= (a – y) (a – 2b + cx – cy)

Question 12.
– 4 (a – 2y)2 + 8 (a – 2y)

Solution:
– 4 (x – 2y)2 + 8 (x – 2y)
= – 4 (x – 2y) (x – 2y) + 8 (x – 2y)
{- 4 (x – 2y) is common}
= – 4 (x – 2y) (x – 2y – 2)
= 4 (x – 2y) (2 – x + 2y)

Question 13.
x3 (a – 2b) + a2 (a – 2b)
Solution:
x3 (a – 2b) + x2 (a – 2b)
HCF of x3, x2 = x2
∴
x2 (a – 2b) (x + 1)
{x2 (x – 2b) is common}
= x2 (x – 2b) (x + 1)

Question 14.
(2x – 3y) (a + b) + (3x – 2y) (a + b)
Solution:
(2x – 3y) (a + b) + (3x – 2y) (a + b)
= (a + b) {2x – 3y + 3x – 2y}
{(x + b) is common}
= (a + b) (5x – 5y)
= 5 (a + b) (x – y)

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

Solution:
4 (x + y) (3a – b) + 6 (a + y) (2b – 3a)
= 4 (x + y) (3a – b) – 6 (x + y) (3a – 2b)
HCF of 4, 6 = 2
= 2 (x + y) {2 (3a – b) – 3 (3a – 2b)}
= 2 (x + 3) {6a – 2b – 9a + 6b}
= 2 (x +y) {-3a + 4b}
= 2 (x + y) (4b – 3a)

All Chapter RD Sharma Solutions For Class 8 Maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 8

All Subject NCERT Solutions For Class 8

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.