Class 8 RD Sharma Solutions – Chapter 7 Factorization – Exercise 7.3

• Difficulty Level : Medium
• Last Updated : 24 Nov, 2020

Question 1. 6x(2x – y) + 7y(2x – y)

Solution:

As we know that (2x – y) is a common factor in the above expression.

So we will take out it and the expression can be written like,

(2x – y)(6x + 7y)

Question 2. 2r(y – x) + s(x – y)

Solution:

We know that

(y – x) = -(x – y) or (x – y) = -(y – x)

So we will write the above expression like

2r(y – x) – s(y – x)

Taking (y – x) as common from this expression

(2r – s)(y – x)

Question 3. 7a(2x – 3) + 3b(2x – 3)

Solution:

We know that in this expression (2x – 3) is common factor,

So we take common from it.

(2x – 3)(7a + 3b)

Question 4. 9a(6a – 5b) – 12a2 (6a – 5b)

Solution:

Taking (6a – 5b) common from the above expression

(9a – 12a2)(6a – 5b)

Now taking a common from it

9a – 12a2

=> 3 * 3a – 4 * 3a

=> 3a(3 – 4a)

Now writing 3a(3 – 4a) instead of (9a – 12a2)

3a(3 – 4a)(6a – 5b)

Question 5. 5(x – 2y)2 + 3(x – 2y)

Solution:

Taking (x – 2y) common from the above expression

(x – 2y)[5(x – 2y) + 3]

=>(x – 2y)(5x – 10y + 3)

Question 6. 16(2l – 3m)2 – 12(3m – 2l)

Solution:

Taking  2l – 3m = -(3m – 2l) or (3m – 2l) = -(2l – 3m)

So above expression can be written as

=> 16(2l – 3m)2 + 12(2l – 3m)

Taking common (2l – 3m)

=> (2l – 3m)[16(2l – 3m) + 12]

=> (2l – 3m){4[4(2l – 3m) + 3}

=> 4(2l – 3m)(8l – 12m + 3)

Question 7. 3a(x – 2y) – b(x – 2y)

Solution:

Taking (x – 2y) common from above expression

=> (x – 2y)(3a – b)

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

Solution:

Taking (x + y) common from above expression

=>(x + y)[ a2 + b2 + c2]

Question 9. (x – y)2 + (x – y)

Solution:

This can be written as

=>(x – y)(x – y) + (x – y)

Taking (x – y) common from the above expression

=> (x – y)[x – y + 1]

Question 10. 6(a + 2b) – 4(a + 2b)2

Solution:

Taking (a + 2b) from the above expression

=> (a + 2b)[6 – 4(a + 2b)]

=> Again taking 2 as a common factor

=> 2(a + 2b)[3 – 2(a + 2b)]

=> 2(a + 2b)(3 – 2a – 4b)

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

Solution:

We can write

y – x = -(x – y)

then

=> a(x – y) – 2b(x – y) + c(x – y)2

Taking (x – y) from above expression

=> (x – y)(a – 2b + c(x – y))

=> (x – y)(a – 2b + cx – cy)

Question 12. -4(x – 2y)2 + 8(x – 2y)

Solution:

Taking 4(x – 2y) as a common factor from expression

=> 4(x – 2y)[-(x – 2y) + 2]

=> 4(x – 2y)[2 – x + 2y]

Question 13. x3(a – 2b) + x2(a – 2b)

Solution:

Taking x2(a – 2b) as common factor from above expression

=> x2(a – 2b)(x + 1)

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

Solution:

Taking (a + b) as a common factor from expression

(a + b)(2x – 3y + 3x – 2y)

(a + b)(5x – 5y)

Taking 5 as common factor from (5x – 5y)

=> 5(a + b)(x – y)

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

Solution:

Taking 2(x + y) as a common factor

=> 2(x + y)[2(3a – b) + 3(2b – 3a)]

=> 2(x + y)[6a – 2b + 6b – 9a]

=> 2(x + y)(4b – 3a)

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