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collins book maths class 8 chapter 7.3 solutions

Answers

Answered by s1048gautham7604
2

Factorize each of the following algebraic expressions:

1. 6x (2x – y) + 7y (2x – y)

Solution:

We have,

6x (2x – y) + 7y (2x – y)

By taking (2x – y) as common we get,

(6x + 7y) (2x – y)

2. 2r (y – x) + s (x – y)

Solution:

We have,

2r (y – x) + s (x – y)

By taking (-1) as common we get,

-2r (x – y) + s (x – y)

By taking (x – y) as common we get,

(x – y) (-2r + s)

(x – y) (s – 2r)

3. 7a (2x – 3) + 3b (2x – 3)

Solution:

We have,

7a (2x – 3) + 3b (2x – 3)

By taking (2x – 3) as common we get,

(7a + 3b) (2x – 3)

4. 9a (6a – 5b) – 12a2 (6a – 5b)

Solution:

We have,

9a (6a – 5b) – 12a2 (6a – 5b)

By taking (6a – 5b) as common we get,

(9a – 12a2) (6a – 5b)

3a(3 – 4a) (6a – 5b)

5. 5 (x – 2y)2 + 3 (x – 2y)

Solution:

We have,

5 (x – 2y)2 + 3 (x – 2y)

By taking (x – 2y) as common we get,

(x – 2y) [5 (x – 2y) + 3]

(x – 2y) (5x – 10y + 3)

6. 16 (2l – 3m)2 – 12 (3m – 2l)

Solution:

We have,

16 (2l – 3m)2 – 12 (3m – 2l)

By taking (-1) as common we get,

16 (2l – 3m)2 + 12 (2l – 3m)

By taking 4(2l – 3m) as common we get,

4(2l – 3m) [4 (2l – 3m) + 3]

4(2l – 3m) (8l – 12m + 3)

7. 3a (x – 2y) – b (x – 2y)

Solution:

We have,

3a (x – 2y) – b (x – 2y)

By taking (x – 2y) as common we get,

(3a – b) (x – 2y)

8. a2 (x + y) + b2 (x + y) + c2 (x + y)

Solution:

We have,

a2 (x + y) + b2 (x + y) + c2 (x + y)

By taking (x + y) as common we get,

(a2 + b2 + c2) (x + y)

9. (x – y)2 + (x – y)

Solution:

We have,

(x – y)2 + (x – y)

By taking (x – y) as common we get,

(x – y) (x – y + 1)

10. 6 (a + 2b) – 4 (a + 2b)2

Solution:

We have,

6 (a + 2b) – 4 (a + 2b)2

By taking (a + 2b) as common we get,

[6 – 4 (a + 2b)] (a + 2b)

(6 – 4a – 8b) (a + 2b)

2(3 – 2a – 4b) (a + 2b)

11. a (x – y) + 2b (y – x) + c (x – y)2

Solution:

We have,

a (x – y) + 2b (y – x) + c (x – y)2

By taking (-1) as common we get,

a (x – y) – 2b (x – y) + c (x – y)2

By taking (x – y) as common we get,

[a – 2b + c(x – y)] (x – y)

(x – y) (a – 2b + cx – cy)

12. -4 (x – 2y)2 + 8 (x – 2y)

Solution:

We have,

-4 (x – 2y)2 + 8 (x – 2y)

By taking 4(x – 2y) as common we get,

[-(x – 2y) + 2] 4(x – 2y)

4(x – 2y) (-x + 2y + 2)

13. x3 (a – 2b) + x2 (a – 2b)

Solution:

We have,

x3 (a – 2b) + x2 (a – 2b)

By taking x2 (a – 2b) as common we get,

(x + 1) [x2 (a – 2b)]

x2 (a – 2b) (x + 1)

14. (2x – 3y) (a + b) + (3x – 2y) (a + b)

Solution:

We have,

(2x – 3y) (a + b) + (3x – 2y) (a + b)

By taking (a + b) as common we get,

(a + b) [(2x – 3y) + (3x – 2y)]

(a + b) [2x -3y + 3x – 2y]

(a + b) [5x – 5y]

(a + b) 5(x – y)

15. 4(x + y) (3a – b) + 6(x + y) (2b – 3a)

Solution:

We have,

4(x + y) (3a – b) + 6(x + y) (2b – 3a)

By taking (x + y) as common we get,

(x + y) [4(3a – b) + 6(2b – 3a)]

(x + y) [12a – 4b + 12b – 18a]

(x + y) [-6a + 8b]

(x + y) 2(-3a + 4b)

(x + y) 2(4b -3a)

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