RD Sharma Solutions for Class 8 Chapter - 7 Factorization Exercise 7.5

RD Sharma Class 8 Solutions Chapter 7 Ex 7.5 PDF Free Download

Get the detailed RD Sharma Solutions for Class 8 Maths Exercise 7.5 Chapter 7 Factorization which is available here. From the exam point of view, subject experts have prepared the solutions in a step by step format in accordance with CBSE syllabus. Students gain knowledge and increase their confidence by following the steps when practised regularly. Download Pdf’s of RD Sharma Solutions from the links, provided below.

“Factorization of binomial expressions expressible as the difference of two squares” is discussed in this Exercise 7.5 of Chapter 7 Factorization.

Download the pdf of RD Sharma For Class 8 Maths Exercise 7.5 Chapter 7 Factorization

 

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rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7
rd sharma class 8 maths chapter 7

Access other Exercises of RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization

Exercise 7.1 Solutions

Exercise 7.2 Solutions

Exercise 7.3 Solutions

Exercise 7.4 Solutions

Exercise 7.6 Solutions

Exercise 7.7 Solutions

Exercise 7.8 Solutions

Exercise 7.9 Solutions

Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 7.5 Chapter 7 Factorization

Factorize each of the following expressions:

1. 16x2 – 25y2

Solution:

We have,

16x2 – 25y2

(4x)2 – (5y)2

By using the formula (a2 – b2) = (a + b) (a – b) we get,

(4x + 5y) (4x – 5y)

2. 27x2 – 12y2

Solution:

We have,

27x2 – 12y2

By taking 3 as common we get,

3 [(3x)2 – (2y)2]

By using the formula (a2 – b2) = (a-b) (a+b)

3 (3x + 2y) (3x – 2y)

3. 144a2 – 289b2

Solution:

We have,

144a2 – 289b2

(12a)2 – (17b)2

By using the formula (a2 – b2) = (a-b) (a+b)

(12a + 17b) (12a – 17b)

4. 12m2 – 27

Solution:

We have,

12m2 – 27

By taking 3 as common we get,

3 (4m2 – 9)

3 [(2m)2 – 32]

By using the formula (a2 – b2) = (a-b) (a+b)

3 (2m + 3) (2m – 3)

5. 125x2 – 45y2

Solution:

We have,

125x2 – 45y2

By taking 5 as common we get,

5 (25x2 – 9y2)

5 [(5x)2 – (3y)2]

By using the formula (a2 – b2) = (a-b) (a+b)

5 (5x + 3y) (5x – 3y)

6. 144a2 – 169b2

Solution:

We have,

144a2 – 169b2

(12a)2 – (13b)2

By using the formula (a2 – b2) = (a-b) (a+b)

(12a + 13b) (12a – 13b)

7. (2a – b)2 – 16c2

Solution:

We have,

(2a – b)2 – 16c2

(2a – b)2 – (4c)2

By using the formula (a2 – b2) = (a-b) (a+b)

(2a – b + 4c) (2a – b – 4c)

8. (x + 2y)2 – 4 (2x – y)2

Solution:

We have,

(x + 2y)2 – 4 (2x – y)2

(x + 2y)2 – [2 (2x – y)]2

By using the formula (a2 – b2)= (a + b) (a – b) we get,

[(x + 2y) + 2 (2x – y)] [x + 2y – 2 (2x – y)]

(x + 4x + 2y – 2y) (x – 4x + 2y + 2y)

(5x) (4y – 3x)

9. 3a5 – 48a3

Solution:

We have,

3a5 – 48a3

By taking 3 as common we get,

3a3 (a2 – 16)

3a3 (a2 – 42)

By using the formula (a2 – b2) = (a-b) (a+b)

3a3 (a + 4) (a – 4)

10. a4 – 16b4

Solution:

We have,

a4 – 16b4

(a2)2 – (4b2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(a2 + 4b2) (a2 – 4b2)

By using the formula (a2 – b2) = (a-b) (a+b)

(a2 + 4b2) (a + 2b) (a – 2b)

11. x8 – 1

Solution:

We have,

x8 – 1

(x4)2–(1)2

By using the formula (a2 – b2) = (a-b) (a+b)

(x4 + 1) (x4 – 1)

By using the formula (a2 – b2) = (a-b) (a+b)

(x4 + 1) (x2 + 1) (x – 1) (x + 1)

12. 64 – (a + 1)2

Solution:

We have,

64 – (a + 1)2

82 – (a + 1)2

By using the formula (a2 – b2) = (a-b) (a+b)

[8 + (a + 1)] [8 – (a + 1)]

(a + 9) (7 – a)

13. 36l2 – (m + n)2

Solution:

We have,

36l2 – (m + n)2

(6l)2 – (m + n)2

By using the formula (a2 – b2) = (a-b) (a+b)

(6l + m + n) (6l – m – n)

14. 25x4y4 – 1

Solution:

We have,

25x4y4 – 1

(5x2y2)2 – (1)2

By using the formula (a2 – b2) = (a-b) (a+b)

(5x2y2 – 1) (5x2y2 + 1)

15. a4 – 1/b4

Solution:

We have,

a4 – 1/b4

(a2)2 – (1/b2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(a2 + 1/b2) (a2 – 1/b2)

By using the formula (a2 – b2) = (a-b) (a+b)

(a2 + 1/b2) (a – 1/b) (a + 1/b)

16. x3 – 144x

Solution:

We have,

x3 – 144x

x [x2 – (12)2]

By using the formula (a2 – b2) = (a-b) (a+b)

x (x + 12) (x – 12)

17. (x – 4y)2 – 625

Solution:

We have,

(x – 4y)2 – 625

(x – 4y)2 – (25)2

By using the formula (a2 – b2) = (a-b) (a+b)

(x – 4y + 25) (x – 4y – 25)

18. 9 (a – b)2 – 100 (x – y)2

Solution:

We have,

9 (a – b)2 – 100 (x – y)2

[3 (a – b)]2 – [10 (x – y)]2

By using the formula (a2 – b2) = (a-b) (a+b)

[3 (a – b) + 10 (x + y)] [3 (a – b) – 10 (x – y)] [3a – 3b + 10x – 10y] [3a – 3b – 10x + 10y]

19. (3 + 2a)2 – 25a2

Solution:

We have,

(3 + 2a)2 – 25a2

(3 + 2a)2 – (5a)2

By using the formula (a2 – b2) = (a-b) (a+b)

(3 + 2a + 5a) (3 + 2a – 5a)

(3 + 7a) (3 – 3a)

(3 + 7a) 3(1 – a)

20. (x + y)2 – (a – b)2

Solution:

We have,

(x + y)2 – (a – b)2

By using the formula (a2 – b2) = (a-b) (a+b)

[(x + y) + (a – b)] [(x + y) – (a – b)]

(x + y + a – b) (x + y – a + b)

21. 1/16x2y2 – 4/49y2z2

Solution:

We have,

1/16x2y2 – 4/49y2z2

(1/4xy)2 – (2/7yz)2

By using the formula (a2 – b2) = (a-b) (a+b)

(xy/4 + 2yz/7) (xy/4 – 2yz/7)

y2 (x/4 + 2/7z) (x/4 – 2/7z)

22. 75a3b2 – 108ab4

Solution:

We have,

75a3b2 – 108ab4

3ab2 (25a2 – 36b2)

3ab2 [(5a)2 – (6b)2]

By using the formula (a2 – b2) = (a-b) (a+b)

3ab2 (5a + 6b) (5a – 6b)

23. x5 – 16x3

Solution:

We have,

x5 – 16x3

x3 (x2 – 16)

x3 (x2 – 42)

By using the formula (a2 – b2) = (a-b) (a+b)

x3 (x + 4) (x – 4)

24. 50/x2 – 2x2/81

Solution:

We have,

50/x2 – 2x2/81

2 (25/x2 – x2/81)

2 [(5/x)2 – (x/9)2]

By using the formula (a2 – b2) = (a-b) (a+b)

2 (5/x+ x/9) (5/x – x/9)

25. 256x3 – 81x

Solution:

We have,

256x3 – 81x

x (256x4 – 81)

x [(16x2)2 – 92]

By using the formula (a2 – b2) = (a-b) (a+b)

x (4x + 3) (4x – 3) (16x2 + 9)

26. a4 – (2b + c)4

Solution:

We have,

a4 – (2b + c)4

(a2)2 – [(2b + c)2]2

By using the formula (a2 – b2) = (a-b) (a+b)

[a2 + (2b + c)2] [a2 – (2b + c)2]

By using the formula (a2 – b2) = (a-b) (a+b)

[a2 + (2b + c)2] [a + 2b + c] [a – 2b – c]

27. (3x + 4y)4 – x4

Solution:

We have,

(3x + 4y)4 – x4

[(3x + 4y)2]2 – (x2)2

By using the formula (a2 – b2) = (a-b) (a+b)

[(3x + 4y)2 + x2] [(3x + 4y)2 – x2] [(3x + 4y)2 + x2] [3x + 4y + x] [3x + 4y – x] [(3x + 4y)2 + x2] [4x + 4y] [2x + 4y] [(3x + 4y)2 + x2] 8[x + 2y] [x + y]

28. p2q2 – p4q4

Solution:

We have,

p2q2 – p4q4

(pq)2 – (p2q2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(pq + p2q2) (pq – p2q2)

p2q2 (1 + pq) (1 – pq)

29. 3x3y – 24xy3

Solution:

We have,

3x3y – 24xy3

3xy (x2 – 8y2)

3xy [x2 – (8y)2]

By using the formula (a2 – b2) = (a-b) (a+b)

(3xy) (x + 8y) (x – 8y)

30. a4b4 – 16c4

Solution:

We have,

a4b4 – 16c4

(a2b2)2 – (4c2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(a2b2 + 4c2) (a2b2 – 4c2)

By using the formula (a2 – b2) = (a-b) (a+b)

(a2b2 + 4c2) (ab + 2c) (ab – 2c)

31. x4 – 625

Solution:

We have,

x4 – 625

(x2)2 – (25)2

By using the formula (a2 – b2) = (a-b) (a+b)

(x2 + 25) (x2 – 25)

(x2 + 25) (x2 – 52)

By using the formula (a2 – b2) = (a-b) (a+b)

(x2 + 25) (x + 5) (x – 5)

32. x4 – 1

Solution:

We have,

x4 – 1

(x2)2 – (1)2

By using the formula (a2 – b2) = (a-b) (a+b)

(x2 + 1) (x2 – 1)

By using the formula (a2 – b2) = (a-b) (a+b)

(x2 + 1) (x + 1) (x – 1)

33. 49(a – b)2 – 25(a + b)2

Solution:

We have,

49(a – b)2 – 25(a + b)2

[7 (a – b)]2 – [5 (a + b)]2

By using the formula (a2 – b2) = (a-b) (a+b)

[7 (a – b) + 5 (a + b)] [7 (a – b) – 5 (a + b)]

(7a – 7b + 5a + 5b) (7a – 7b – 5a – 5b)

(12a – 2b) (2a – 12b)

2 (6a – b) 2 (a – 6b)

4 (6a – b) (a – 6b)

34. x – y – x2 + y2

Solution:

We have,

x – y – x2 + y2

x – y – (x2 – y2)

By using the formula (a2 – b2) = (a-b) (a+b)

x – y – (x + y) (x – y)

(x – y) (1 – x – y)

35. 16(2x – 1)2 – 25y2

Solution:

We have,

16(2x – 1)2 – 25y2

[4 (2x – 1)]2 – (5y)2

By using the formula (a2 – b2) = (a-b) (a+b)

(8x + 5y – 4) (8x – 5y – 4)

36. 4(xy + 1)2 – 9(x – 1)2

Solution:

We have,

4(xy + 1)2 – 9(x – 1)2

[2x (xy + 1)]2 – [3 (x – 1)]2

By using the formula (a2 – b2) = (a-b) (a+b)

(2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)

(2xy + 3x – 1) (2xy – 3x + 5)

37. (2x + 1)2 – 9x4

Solution:

We have,

(2x + 1)2 – 9x4

(2x + 1)2 – (3x2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(2x + 1 + 3x2) (2x + 1 – 3x2)

(3x2 + 2x + 1) (-3x2 + 2x + 1)

38. x4 – (2y – 3z)2

Solution:

We have,

x4 – (2y – 3z)2

(x2)2 – (2y – 3z)2

By using the formula (a2 – b2) = (a-b) (a+b)

(x2 + 2y – 3z) (x2 – 2y + 3z)

39. a4 – b2 + a – b

Solution:

We have,

a4 – b2 + a – b

By using the formula (a2 – b2) = (a-b) (a+b)

(a + b) (a – b) + (a – b)

(a – b) (a + b + 1)

40. 16a4 – b4

Solution:

We have,

16a4 – b4

(4a2)2 – (b2)2

(4a2 + b2) (4a2 – b2)

By using the formula (a2 – b2) = (a-b) (a+b)

(4a2 + b2) (2a + b) (2a – b)

41. a4 – 16(b – c)4

Solution:

We have,

a4 – 16(b – c)4

(a2)2 – [4 (b – c)2]

By using the formula (a2 – b2) = (a-b) (a+b)

[a2 + 4 (b – c)2] [a2 – 4 (b – c)2]

By using the formula (a2 – b2) = (a-b) (a+b)

[a2 + 4 (b – c)2] [(a + 2b – 2c) (a – 2b + 2c)]

42. 2a4 – 32a

Solution:

We have,

2a4 – 32a

2a (a4 – 16)

2a [(a)2 – (4)2]

By using the formula (a2 – b2) = (a-b) (a+b)

2a (a2 + 4) (a2 – 4)

2a (a2 + 4) (a2 – 22)

By using the formula (a2 – b2) = (a-b) (a+b)

2a (a2 + 4) (a + 2) (a – 2)

43. a4b4 – 81c4

Solution:

We have,

a4b4 – 81c4

(a2b2)2 – (9c2)2

By using the formula (a2 – b2) = (a-b) (a+b)

(a2b2 + 9c2) (a2b2 – 9c2)

By using the formula (a2 – b2) = (a-b) (a+b)

(a2b2 + 9c2) (ab + 3c) (ab – 3c)

44. xy9 – yx9

Solution:

We have,

xy9 – yx9

-xy (x8 – y8)

-xy [(x4)2 – (y4)2]

By using the formula (a2 – b2) = (a-b) (a+b)

-xy (x4 + y4) (x4 – y4)

By using the formula (a2 – b2) = (a-b) (a+b)

-xy (x4 + y4) (x2 + y2) (x2 – y2)

By using the formula (a2 – b2) = (a-b) (a+b)

-xy (x4 + y4) (x2 + y2) (x + y) (x – y)

45. x3 – x

Solution:

We have,

x3 – x

x (x2 – 1)

By using the formula (a2 – b2) = (a-b) (a+b)

x (x + 1) (x – 1)

46. 18a2x2 – 32

Solution:

We have,

18a2x2 – 32

2 [(3ax)2 – (4)2]

By using the formula (a2 – b2) = (a-b) (a+b)

2 (3ax + 4) (3ax – 4)


 

1 Comment

  1. Great solutions
    Thank you so much

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