 # RD Sharma Solutions for Class 8 Maths Chapter - 7 Factorization Exercise 7.4

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This Exercise 7.4 of Chapter 7 Factorization is on based on the factorization by grouping the terms.

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#### EXERCISE 7.4 PAGE NO: 7.12

Factorize each of the following expressions:

1. qr – pr + qs – ps

Solution:

We have,

qr – pr + qs – ps

By grouping similar terms we get,

qr + qs – pr – ps

q(r + s) –p (r + s)

(q – p) (r + s)

2. p2q – pr2 – pq + r2

Solution:

We have,

p2q – pr2 – pq + r2

By grouping similar terms we get,

p2q – pq – pr2 + r2

pq(p – 1) –r2(p – 1)

(p – 1) (pq – r2)

3. 1 + x + xy + x2y

Solution:

We have,

1 + x + xy + x2y

1 (1 + x) + xy(1 + x)

(1 + x) (1 + xy)

4. ax + ay – bx – by

Solution:

We have,

ax + ay – bx – by

a(x + y) –b (x + y)

(a – b) (x + y)

5. xa2 + xb2 – ya2 – yb2

Solution:

We have,

xa2 + xb2 – ya2 – yb2

x(a2 + b2) –y (a2 + b2)

(x – y) (a2 + b2)

6. x2 + xy + xz + yz

Solution:

We have,

x2 + xy + xz + yz

x (x + y) + z (x + y)

(x + y) (x + z)

7. 2ax + bx + 2ay + by

Solution:

We have,

2ax + bx + 2ay + by

By grouping similar terms we get,

2ax + 2ay + bx + by

2a (x + y) + b (x + y)

(2a + b) (x + y)

8. ab – by – ay + y2

Solution:

We have,

ab – by – ay + y2

By grouping similar terms we get,

Ab – ay – by + y2

a (b – y) – y (b – y)

(a – y) (b – y)

9. axy + bcxy – az – bcz

Solution:

We have,

axy + bcxy – az – bcz

By grouping similar terms we get,

axy – az + bcxy – bcz

a (xy – z) + bc (xy – z)

(a + bc) (xy – z)

10. lm2 – mn2 – lm + n2

Solution:

We have,

lm2 – mn2 – lm + n2

By grouping similar terms we get,

lm2 – lm – mn2 + n2

lm (m – 1) – n2 (m – 1)

(lm – n2) (m – 1)

11. x3 – y2 + x – x2y2

Solution:

We have,

x3 – y2 + x – x2y2

By grouping similar terms we get,

x + x3 – y2 – x2y2

x (1 + x2) – y2 (1 + x2)

(x – y2) (1 + x2)

12. 6xy + 6 – 9y – 4x

Solution:

We have,

6xy + 6 – 9y – 4x

By grouping similar terms we get,

6xy – 4x – 9y + 6

2x (3y – 2) – 3 (3y – 2)

(2x – 3) (3y – 2)

13. x2 – 2ax – 2ab + bx

Solution:

We have,

x2 – 2ax – 2ab + bx

By grouping similar terms we get,

x2 + bx – 2ax – 2ab

x (x + b) – 2a (x + b)

(x – 2a) (x + b)

14. x3 – 2x2y + 3xy2 – 6y3

Solution:

We have,

x3 – 2x2y + 3xy2 – 6y3

By grouping similar terms we get,

x3 + 3xy2 – 2x2y – 6y3

x (x2 + 3y2) – 2y (x2 + 3y2)

(x – 2y) (x2 + 3y2)

15. abx2 + (ay – b) x – y

Solution:

We have,

abx2 + (ay – b) x – y

abx2 + ayx – bx – y

By grouping similar terms we get,

abx2 – bx + ayx – y

bx (ax – 1) + y (ax – 1)

(bx + y) (ax – 1)

16. (ax + by)2 + (bx – ay)2

Solution:

We have,

(ax + by)2 + (bx – ay)2

a2x2 + b2y2 + 2axby + b2x2 + a2y2 – 2axby

a2x2 + b2y2 + b2x2 + a2y2

By grouping similar terms we get,

a2x2 + a2y2 + b2y2 + b2x2

a2 (x2 + y2) + b2 (x2 + y2)

(a2 + b2) (x2 + y2)

17. 16 (a – b)3 – 24 (a – b)2

Solution:

We have,

16(a – b)3 – 24(a – b)2

8 (a – b)2 [2 (a – b) – 3]

8 (a – b)2 (2a – 2b – 3)

18. ab (x2 + 1) + x (a2 + b2)

Solution:

We have,

ab(x2 + 1) + x(a2 + b2)

abx2 + ab + xa2 + xb2

By grouping similar terms we get,

abx2 + xa2 + xb2 + ab

ax (bx + a) + b (bx + a)

(ax + b) (bx + a)

19. a2x2 + (ax2 + 1)x + a

Solution:

We have,

a2x2 + (ax2 + 1)x + a

a2x2 + ax3 + x + a

ax2 (a + x) + 1 (x + a)

(x + a) (ax2 + 1)

20. a (a – 2b – c) + 2bc

Solution:

We have,

a (a – 2b – c) + 2bc

a2 – 2ab – ac + 2bc

a (a – 2b) – c (a – 2b)

(a – 2b) (a – c)

21. a (a + b – c) – bc

Solution:

We have,

a (a + b – c) – bc

a2 + ab – ac – bc

a (a + b) – c (a + b)

(a + b) (a – c)

22. x2 – 11xy – x + 11y

Solution:

We have,

x2 – 11xy – x + 11y

By grouping similar terms we get,

x2 – x – 11xy + 11y

x (x – 1) – 11y (x – 1)

(x – 11y) (x – 1)

23. ab – a – b + 1

Solution:

We have,

ab – a – b + 1

a (b – 1) – 1 (b – 1)

(a – 1) (b – 1)

24. x2 + y – xy – x

Solution:

We have,

x2 + y – xy – x

By grouping similar terms we get,

x2 – x + y – xy

x (x – 1) – y (x – 1)

(x – y) (x – 1)