Students can download the free PDF of RD Sharma Solutions for Class 8 Maths Exercise 7.4 Chapter 7 Factorization, which is provided here. The solutions are designed by the expert faculty team at BYJU’S, with utmost care, to help students understand the concepts clearly. Students are suggested to practise these solutions on a regular basis, which helps them in gaining knowledge of all the concepts, and thus attain good marks in their exams. The PDFs of RD Sharma Solutions can be downloaded easily from the links given below.
Exercise 7.4 of Chapter 7 Factorization is based on factorization by grouping the terms.
RD Sharma Solutions for Class 8 Maths Exercise 7.4 Chapter 7 Factorization
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.4 Chapter 7 Factorization
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)
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