collins book maths class 8 chapter 7.3 solutions
Answers
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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