Exercise 7.6 explains the topic factorization of algebraic expressions expressible as a perfect square. Students can get the detailed RD Sharma Solutions for Class 8 Maths Exercise 7.6 Chapter 7 Factorization from the links available below. Students will gain better knowledge about the exercise-wise problems in this chapter by solving RD Sharma Solutions. To score well in the final exams, students are advised to regularly practise the RD Sharma Solutions provided by the experts at BYJU’S. PDFs can be downloaded easily from the links given below.
RD Sharma Solutions for Class 8 Maths Exercise 7.6 Chapter 7 Factorization
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.6 Chapter 7 Factorization
EXERCISE 7.6 PAGE NO: 7.22
Factorize each of the following algebraic expressions:
1. 4x2 + 12xy + 9y2
Solution:
We have,
4x2 + 12xy + 9y2
By using the formula (x + y)2 = x2 + y2 + 2xy
(2x)2 + (3y)2 + 2 (2x) (3y)
(2x + 3y)2
(2x + 3y) (2x + 3y)
2. 9a2 – 24ab + 16b2
Solution:
We have,
9a2 – 24ab + 16b2
By using the formula (x – y)2 = x2 + y2 – 2xy
Here x = 3a, y = 4b So,
(3a)2 + (4b)2 – 2 (3a) (4b)
(3a – 4b)2
(3a – 4b) (3a – 4b)
3. p2q2 – 6pqr + 9r2
Solution:
We have,
p2q2 – 6pqr + 9r2
By using the formula (a – b)2 = a2 + b2 – 2ab
(pq)2 + (3r)2 – 2 (pq) (3r)
(pq – 3r)2
(pq – 3r) (pq – 3r)
4. 36a2 + 36a + 9
Solution:
We have,
36a2 + 36a + 9
(6a)2 + 2 × 6a × 3 + 32
(6a + 3)2
5. a2 + 2ab + b2 – 16
Solution:
We have,
a2 + 2ab + b2 – 16
By using the formula (a – b)2 = a2 + b2 – 2ab
(a + b)2 – 42
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b + 4) (a + b – 4)
6. 9z2 – x2 + 4xy – 4y2
Solution:
We have,
9z2 – x2 + 4xy – 4y2
(3z)2 – [x2 – 2 (x) (2y) + (2y)2]
By using the formula (a – b)2 = a2 + b2 – 2ab
(3z)2 – (x – 2y)2
By using the formula (a2 – b2) = (a+b) (a-b)
[(x – 2y) + 3z] [–x + 2y + 3z)]7. 9a4 – 24a2b2 + 16b4 – 256
Solution:
We have,
9a4 – 24a2b2 + 16b4 – 256
(3a2)2 – 2 (4a2) (3b2) + (4b2)2 – (16)2
By using the formula (a – b)2 = a2 + b2 – 2ab
(3a2 – 4b2)2 – (16)2
By using the formula (a2 – b2) = (a+b) (a-b)
(3a2 – 4b2 + 16) (3a2 – 4b2 – 16)
8. 16 – a6 + 4a3b3 – 4b6
Solution:
We have,
16 – a6 + 4a3b3 – 4b6
42 – [(a3)2 – 2 (a3) (2b3) + (2b3)2]
By using the formula (a – b)2 = a2 + b2 – 2ab
42 – (a3 – 2b3)2
By using the formula (a2 – b2) = (a+b) (a-b)
[4 + (a3 – 2b3)] [4 – (a3 – 2b3)]9. a2 – 2ab + b2 – c2
Solution:
We have,
a2 – 2ab + b2 – c2
By using the formula (a – b)2 = a2 + b2 – 2ab
(a – b)2 – c2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – b + c) (a – b – c)
10. x2 + 2x + 1 – 9y2
Solution:
We have,
x2 + 2x + 1 – 9y2
By using the formula (a – b)2 = a2 + b2 – 2ab
(x + 1)2 – (3y)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + 3y + 1) (x – 3y + 1)
11. a2 + 4ab + 3b2
Solution:
We have,
a2 + 4ab + 3b2
By using factors for 3 i.e., 3 and 1
a2 + ab + 3ab + 3b2
By grouping, we get,
a (a + b) + 3b (a + b)
(a + 3b) (a + b)
12. 96 – 4x – x2
Solution:
We have,
96 – 4x – x2
-x2 – 4x + 96
By using factors for 96 i.e., 12 and 8
-x2 – 12x + 8x + 96
By grouping, we get,
-x (x + 12) + 8 (x + 12)
(x + 12) (-x + 8)
13. a4 + 3a2 + 4
Solution:
We have,
a4 + 3a2 + 4
(a2)2 + (a2)2 + 2 (2a2) + 4 – a2
(a2 + 2)2 + (-a2)
By using the formula (a2 – b2) = (a+b) (a-b)
(a2 + 2 + a) (a2 + 2 – a)
(a2 + a + 2) (a2 – a + 2)
14. 4x4 + 1
Solution:
We have,
4x4 + 1
(2x2)2 + 1 + 4x2 – 4x2
(2x2 + 1)2 – 4x2
By using the formula (a2 – b2) = (a+b) (a-b)
(2x2 + 1 + 2x) (2x2 + 1 – 2x)
(2x2 + 2x + 1) (2x2 – 2x + 1)
15. 4x4 + y4
Solution:
We have,
4x4 + y4
(2x2)2 + (y2)2 + 4x2y2 – 4x2y2
(2x2 + y2)2 – 4x2y2
By using the formula (a2 – b2) = (a+b) (a-b)
(2x2 + y2 + 2xy) (2x2 + y2 – 2xy)
16. (x + 2)4 – 6(x + 2) + 9
Solution:
We have,
(x + 2)4 – 6(x + 2) + 9
(x2 + 22)2 – 6x – 12 + 9
(x2 + 22 + 2(2)(x)) – 6x – 12 + 9
x2 + 4 + 4x – 6x – 12 + 9
x2 – 2x + 1
By using the formula (a – b)2 = a2 + b2 – 2ab
(x – 1)2
17. 25 – p2 – q2 – 2pq
Solution:
We have,
25 – p2 – q2 – 2pq
25 – (p2 + q2 + 2pq)
(5)2 – (p + q)2
By using the formula (a2 – b2) = (a+b) (a-b)
(5 + p + q) (5 –p – q)
-(p + q + 5) (p + q – 5)
18. x2 + 9y2 – 6xy – 25a2
Solution:
We have,
x2 + 9y2 – 6xy – 25a2
(x – 3y)2 – (5a)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x – 3y + 5a) (x – 3y – 5a)
19. 49 – a2 + 8ab – 16b2
Solution:
We have,
49 – a2 + 8ab – 16b2
49 – (a2 – 8ab + 16b2)
49 – (a – 4b)2
By using the formula (a2 – b2) = (a + b) (a – b)
(7 + a – 4b) (7 – a + 4b)
-(a – 4b + 7) (a – 4b – 7)
20. a2 – 8ab + 16b2 – 25c2
Solution:
We have,
a2 – 8ab + 16b2 – 25c2
(a – 4b)2– (5c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – 4b + 5c) (a – 4b – 5c)
21. x2 – y2 + 6y – 9
Solution:
We have,
x2 – y2 + 6y – 9
x2 + 6y – (y2 – 6y + 9)
x2 – (y – 3)2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + y – 3) (x – y + 3)
22. 25x2 – 10x + 1 – 36y2
Solution:
We have,
25x2 – 10x + 1 – 36y2
(5x)2 – 2 (5x) + 1 – (6y)2
(5x – 1)2 – (6y)2
By using the formula (a2 – b2) = (a+b) (a-b)
(5x – 6y – 1) (5x + 6y – 1)
23. a2 – b2 + 2bc – c2
Solution:
We have,
a2 – b2 + 2bc – c2
a2 – (b2 – 2bc + c2)
a2 – (b – c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b – c) (a – b + c)
24. a2 + 2ab + b2 – c2
Solution:
We have,
a2 + 2ab + b2 – c2
(a + b)2 – c2
By using the formula (a2 – b2) = (a+b) (a-b)
(a + b + c) (a + b – c)
25. 49 – x2 – y2 + 2xy
Solution:
We have,
49 – x2 – y2 + 2xy
49 – (x2 + y2 – 2xy)
72 – (x – y)2
By using the formula (a2 – b2) = (a+b) (a-b)
[7 + (x – y)] [7 – x + y](x – y + 7) (y – x + 7)
26. a2 + 4b2 – 4ab – 4c2
Solution:
We have,
a2 + 4b2 – 4ab – 4c2
a2 – 2 (a) (2b) + (2b)2 – (2c)2
(a – 2b)2 – (2c)2
By using the formula (a2 – b2) = (a+b) (a-b)
(a – 2b + 2c) (a – 2b – 2c)
27. x2 – y2 – 4xz + 4z2
Solution:
We have,
x2 – y2 – 4xz + 4z2
x2 – 2 (x) (2z) + (2z)2 – y2
As (a-b)2 = a2 + b2 – 2ab
(x – 2z)2 – y2
By using the formula (a2 – b2) = (a+b) (a-b)
(x + y – 2z) (x – y – 2z)
Comments