RD Sharma Solutions Class 8 Factorization Exercise 7.3

RD Sharma Solutions Class 8 Chapter 7 Exercise 7.3

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

Factorize each of the following algebraic expressions :

Q.1)  6x(2x – y) + 7y(2x – y)

Soln.:

6x(2x – y) + 7y(2x – y)

= (6x + 7y)(2x – y) (taking (2x – y) as common factor)

Q.2)  2r(y – x) + s(x – y) 

Soln.:

2r(y – x) + s(x – y)

=  2r(y – x) – s(y – x) [since, (x – y) = -(y – x)]

= (2r – s)(y – x) [taking (y – x) as the common factor]

Q.3)  7a(2x – 3) + 3b(2x – 3)

Soln.:

7a(2x – 3) + 3b(2x – 3)

= (7a + 3b)(2x – 3) [taking (2x – 3) as the common factor]

Q.4)  9a(6a – 5b) – 12a2(6a – 5b)

Soln.:

9a(6a – 5b) – 12a2(6a – 5b)

= (9a – 12qa2)(6a – 5b) [taking (6a – 5b) as the common factor]

= 3a(3 – 4a)(6a – 5b) [taking 3a as the common factor of the quadratic eqn. (9a – 12a2)]

Q.5)  5(x – 2y)2 + 3(x – 2y)

Soln.:

5(x – 2y)2 + 3(x – 2y)

= [(x – 2y) + 3](x – 2y) [taking (x – 2y) as the common factor]

= (5x – 10y + 3)(x – 2y)

Q.6)  16(2L – 3m)2 – 12(3m – 2L)

Soln.:

16(2L – 3m)2 – 12(3m – 2L)

= 16(2L – 3m)2 + 12(2L – 3m) [(3m – 2L) =  -(2L – 3m)]

= [16(2L – 3m) + 12](2L – 3m) [taking (2L – 3m) as the common factor]

= 4[4(2L – 3m) + 3](2L – 3m) [taking 4 as the common factor (16(2L – 3m) + 12)]

= 4(8L – 12m + 3)(2L – 3m)

Q.7)  3a(x – 2y) – b(x – 2y)

Soln.:

3a(x – 2y) – b(x – 2y)

= (3a  -b)(x – 2y) [taking (x – 2y) as the common factor]

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

Soln.:

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

= (a2 + b2 + c2)(x + y) [taking (x +y) as the common the factor]

Q.9)  (x – y)2 + (x – y)

Soln.:

(x – y)2 + (x – y)

= (x – y)(x – y) + (x – y) [taking (x – y) as the common factor]

= (x – y + 1)(x – y)

Q.10)  6(a + 2b) – 4(a +2b)2

Soln.:

6(a + 2b) – 4(a +2b)2

= [6 – 4(a + 2b)](a + 2b) [taking (a + 2b as the common factor)]

= 2[3 – 2(a + 2b)](a + 2b) [taking 2 as the common factor of [6 – 4(a + 2b)]]

= 2(3 – 2a – 4b)(a + 2b)

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

Soln.:

a(x – y) + 2b(y – x) + c(x – y)2

= a(x – y) – 2b(x -y) +c(x – y)2 [(y -x) = -(x – y)]

= [a – 2b + c(x- y)](x – y)

= (a – 2b + cx – cy)(x- y)

Q.12)  -4(x – 2y)2 + 8(x – 2y)

Soln.:

-4(x – 2y)2 + 8(x – 2y)

= [-4(x – 2y) + 8](x -2y) [taking (x – 2y) as the common factor]

= 4[-(x – 2y) + 2](x – 2y) [taking 4 as the common factor of [-4(x – 2y) + 8]]

= 4(2y – x + 2)(x – 2y)

Q.13)  x3(a – 2b) + x2(a – 2b)

Soln.:

x3(a – 2b) + x2(a – 2b)

= (x3 + x2)(a – 2b) [taking (a – 2b) as the common factor]

= x2(x + 1)(a – 2b) [taking x2 as the common factor of (x3 + x2)]

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

Soln.:

(2x – 3y)(a + b) + (3x – 2y)(a + b)

= (2x – 3y + 3x – 2y)(a +b) [taking (a +b) as the common factor]

= (5x – 5y)(a + b)

= 5(x – y)(a + b) [taking 5 as the common factor of (5x – 5y)]

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

Soln.:

4(x + y)(3a – b) + 6(x + y)(2b – 3a)

= 2(x + y)[2(3a – b) + 3(2b – 3a)] [taking (2(x + y)) as the common factor ]

= 2(x + y)(6a – 2b + 6b – 9a)

= 2(x + y)(4b – 3a)