Chapter 7: Factorization Exercise – 7.3

Question: 1

Factorize of the following algebraic expressions:

6x(2x – y) + 7y(2x – y)

Solution:

6x(2x – y) + 7y(2x – y)

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

 

Question: 2

Factorize of the following algebraic expressions:

2r(y – x) + s(x – y)

Solution:

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]

 

Question: 3

Factorize of the following algebraic expressions:

7a(2x – 3) + 3b(2x – 3)

Solution:

7a(2x – 3) + 3b(2x – 3)

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

 

Question: 4

Factorize of the following algebraic expressions:

9a(6a – 5b) – 12a²(6a – 5b)

Solution:

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)]

 

Question: 5

Factorize of the following algebraic expressions:

5(x – 2y)² + 3(x – 2y)

Solution:

5(x – 2y)² + 3(x – 2y)

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

= (5x – 10y + 3)(x – 2y)

 

Question: 6

Factorize of the following algebraic expressions:

16(2L – 3m)² -12(3m - 2L)

Solution:

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)

 

Question: 7

Factorize of the following algebraic expressions:

3a(x – 2y) – b(x – 2y)

Solution:

3a(x – 2y) – b(x – 2y)

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

 

Question: 8

Factorize of the following algebraic expressions:

a²(x + y) + b²(x + y) + c²(x + y)

Solution:

a²(x + y) + b²(x + y) +c²(x + y)

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

 

Question: 9

Factorize of the following algebraic expressions:

(x – y)² + (x – y)

Solution:

(x – y)² + (x – y)

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

= (x – y + 1)(x – y)

 

Question: 10

Factorize of the following algebraic expressions:

6(a + 2b) – 4(a +2b)²

Solution:

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)

 

Question: 11

Factorize of the following algebraic expressions:

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)² [(y -x) = -(x – y)]

= [a – 2b + c(x- y)](x – y)

= (a – 2b + cx – cy)(x- y)

 

Question: 12

Factorize of the following algebraic expressions:

- 4(x – 2y)² + 8(x – 2y)

Solution:

-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)

 

Question: 13

Factorize of the following algebraic expressions:

x³(a – 2b) + x²(a – 2b)

Solution:

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)]

 

Question: 14

Factorize of the following algebraic expressions:

(2x – 3y)(a + b) + (3x – 2y)(a + b)

Solution:

(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)]

 

Question: 15

Factorize of the following algebraic expressions:

4(x + y)(3a – b) + 6(x + y)(2b – 3a)

Solution:

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)