Exercise 3C RS Aggarwal Class 9 Factorisation of Polynomials Best Solutions

Exercise 3C RS Aggarwal Class 9 Mathematics Solutions

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The question number 1 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

Factorise:

1. $x^{2} + 11 x + 30$

Answer

We have
$x^{2} + 11 x + 30$

$= x^{2} + (6 + 5) x + 30$ ... $(∵ 30 = 6 \times 5, 11 = 6 + 5)$

$= x^{2} + 6 x + 5 x + 30$

$= x (x + 6) + 5 (x + 6)$

$= (x + 6) (x + 5)$ .

The question number 2 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

2. $x^{2} + 18 x + 32$

Answer

We have
$x^{2} + 18 x + 32$

$= x^{2} + (16 + 2) x + 32$ ... $(∵ 32 = 16 \times 2, 18 = 16 + 2)$

$= x^{2} + 16 x + 2 x + 32$

$= x (x + 16) + 2 (x + 16)$

$= (x + 16) (x + 2)$

The question number 3 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

3. $x^{2} + 20 x - 69$

Answer

We have
$x^{2} + 20 x - 69$

$= x^{2} + (23 - 3) x - 69$ ... $[∵ - 69 = 23 \times (- 3), 20 = 23 - 3]$

$= x^{2} + 23 x - 3 x - 69$

$= x (x + 23) - 2 (x + 23)$

$= (x + 23) (x - 2)$

The question number 4 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

4. $x^{2} + 19 x - 150$

Answer

We have
$x^{2} + 19 x - 150$

Here, $150 = 2 \times 3 \times 5 \times 5 \times 1$
$\Rightarrow 150 = (2 \times 3) \times (5 \times 5 \times 1)$
$\Rightarrow 150 = 6 \times 25$

Now,
$x^{2} + 19 x - 150$

$= x^{2} + (25 - 6) x - 150$ ... $[∵ - 150 = 25 \times (- 6), 19 = 25 - 6]$

$= x^{2} + 25 x - 6 x - 150$

$= x (x + 25) - 6 (x + 25)$

$= (x + 25) (x - 6)$ .

The question number 5 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

5. $x^{2} + 7 x - 98$

Answer

We have
$x^{2} + 7 x - 98$

Here, $98 = 2 \times 7 \times 7 \times 1$
$\Rightarrow 98 = (2 \times 7) \times (7 \times 1)$
$\Rightarrow 98 = 14 \times 7$

Now,
$x^{2} + 7 x - 98$

$= x^{2} + (14 - 7) x - 98$ ... $[∵ - 98 = 14 \times (- 7), 7 = 14 - 7]$

$= x^{2} + 14 x - 7 x - 98$

$= x (x + 14) - 7 (x + 14)$

$= (x + 14) (x - 7)$ .

The question number 6 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

6. $x^{2} + 2 \sqrt{3} x - 24$

Answer

We have
$x^{2} + 2 \sqrt{3} x - 24$

Here, $24 = 2 \times 2 \times 3 \times 2 \times 1$
$\Rightarrow 24 = 2 \times 2 \times \sqrt{3} \times \sqrt{3} \times 2 \times 1$
$\Rightarrow 24 = (2 \times 2 \times \sqrt{3}) \times (\sqrt{3} \times 2 \times 1)$
$\Rightarrow 24 = 4 \sqrt{3} \times 2 \sqrt{3}$

Now,
$x^{2} + 2 \sqrt{3} x - 24$

$= x^{2} + (4 \sqrt{3} - 2 \sqrt{3}) x - 24$ ... $[∵ - 24 = 4 \sqrt{3} \times (- 2 \sqrt{3}), 2 \sqrt{3} = 4 \sqrt{3} - 2 \sqrt{3}]$

$= x^{2} + 4 \sqrt{3} x - 2 \sqrt{3} x - 24$

$= x (x + 4 \sqrt{3}) - 2 \sqrt{3} (x + 4 \sqrt{3})$

$= (x + 4 \sqrt{3}) (x - 2 \sqrt{3})$ .

The question number 7 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

7. $x^{2} - 21 x + 90$

Answer

We have
$x^{2} - 21 x + 90$

Here, $90 = 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 90 = (2 \times 3) \times (3 \times 5 \times 1)$
$\Rightarrow 90 = 6 \times 15$

Now,
$x^{2} - 21 x + 90$

$= x^{2} + {(- 15) + (- 6)} x + 90$ ... $[∵ 90 = (- 15) \times (- 6), - 21 = (- 15) + (- 6)]$

$= x^{2} + (- 15) x + (- 6) x + 90$

$= x^{2} - 15 x - 6 x + 90$

$= x (x - 15) - 6 (x - 15)$

$= (x - 15) (x - 6)$ .

The question number 8 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

8. $x^{2} - 22 x + 120$

Answer

We have
$x^{2} - 22 x + 120$

Here, $120 = 2 \times 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 120 = (2 \times 2 \times 3) \times (2 \times 5 \times 1)$
$\Rightarrow 120 = 12 \times 10$

Now,
$x^{2} - 22 x + 120$

$= x^{2} + {(- 12) + (- 10)} x + 120$ ... $[∵ 120 = (- 12) \times (- 10), - 22 = (- 12) + (- 10)]$

$= x^{2} + (- 12) x + (- 10) x + 120$

$= x^{2} - 12 x - 10 x + 120$

$= x (x - 12) - 10 (x - 15)$

$= (x - 12) (x - 15)$

The question number 9 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

9. $x^{2} - 4 x + 3$

Answer

We have
$x^{2} - 4 x + 3$

$= x^{2} + {(- 3) + (- 1)} x + 3$ ... $[∵ 3 = (- 3) \times (- 1), - 4 = (- 3) + (- 1)]$

$= x^{2} - 3 x - x + 3$

$= x (x - 3) - 1 (x - 3)$

$= (x - 3) (x - 1)$

The question number 10 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

10. $x^{2} + 7 \sqrt{6} x + 60$

Answer

We have
$x^{2} + 7 \sqrt{6} x + 60$

Here, $60 = 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 60 = 2 \times 6 \times 5 \times 1$
$\Rightarrow 60 = 2 \times \sqrt{6} \times \sqrt{6} \times 5 \times 1$
$\Rightarrow 60 = (2 \times \sqrt{6}) \times (\sqrt{6} \times 5 \times 1)$
$\Rightarrow 60 = 2 \sqrt{6} \times 5 \sqrt{6}$

Now,
$x^{2} + 7 \sqrt{6} x + 60$

$= x^{2} + (5 \sqrt{6} + 2 \sqrt{6}) x + 60$ ... $[∵ 60 = 5 \sqrt{6} \times 2 \sqrt{6}, 7 \sqrt{6} = 5 \sqrt{6} + 2 \sqrt{6}]$

$= x^{2} + 5 \sqrt{6} x + 2 \sqrt{6} x + 60$

$= x (x + 5 \sqrt{6}) + 2 \sqrt{6} (x + 5 \sqrt{6})$

$= (x + 5 \sqrt{6}) (x + 2 \sqrt{6})$

The question number 11 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

11. $x^{2} + 3 \sqrt{3} x + 6$

Answer

We have
$x^{2} + 3 \sqrt{3} x + 6$

Here, $6 = 2 \times 3 \times 1$
$\Rightarrow 6 = 2 \times \sqrt{3} \times \sqrt{3} \times 1$
$\Rightarrow 6 = (2 \times \sqrt{3}) \times (\sqrt{3} \times 1)$
$\Rightarrow 6 = 2 \sqrt{3} \times \sqrt{3}$

Now,
$x^{2} + 3 \sqrt{3} x + 6$

$= x^{2} + (2 \sqrt{3} + \sqrt{3}) x + 6$ ... $[∵ 6 = 2 \sqrt{3} \times \sqrt{3}, 3 \sqrt{3} = 2 \sqrt{3} + \sqrt{3}]$

$= x^{2} + 2 \sqrt{3} x + \sqrt{3} x + 6$

$= x (x + 2 \sqrt{3}) + \sqrt{3} (x + 2 \sqrt{3})$

$= (x + 2 \sqrt{3}) (x + \sqrt{3})$

The question number 12 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

12. $x^{2} + 6 \sqrt{6} x + 48$

Answer

We have
$x^{2} + 6 \sqrt{6} x + 48$

Here,
$48 = 2 \times 2 \times 2 \times 2 \times 3 \times 1$
$\Rightarrow 48 = 2 \times 2 \times 2 \times 6 \times 1$
$\Rightarrow 48 = 2 \times 2 \times 2 \times \sqrt{6} \times \sqrt{6} \times 1$
$\Rightarrow 48 = 2 \times 2 \times \sqrt{6} \times 2 \times \sqrt{6} \times 1$
$\Rightarrow 48 = (2 \times 2 \times \sqrt{6}) \times (2 \times \sqrt{6} \times 1)$
$\Rightarrow 48 = 4 \sqrt{6} \times 2 \sqrt{6}$

Now,
$x^{2} + 6 \sqrt{6} x + 48$

$= x^{2} + (4 \sqrt{6} + 2 \sqrt{6}) x + 48$

$= x^{2} + 4 \sqrt{6} x + 2 \sqrt{6} x + 48$

$= x (x + 4 \sqrt{6}) + 2 \sqrt{6} (x + 4 \sqrt{6})$

$= (x + 4 \sqrt{6}) (x + 2 \sqrt{6})$

The question number 13 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

13. $x^{2} + 5 \sqrt{5} x + 30$

Answer

We have
$x^{2} + 5 \sqrt{5} x + 30$

Here,
$30 = 2 \times 3 \times 5 \times 1$
$\Rightarrow 30 = 2 \times 3 \times \sqrt{5} \times \sqrt{5} \times 1$
$\Rightarrow 30 = (2 \times \sqrt{5}) \times (3 \times \sqrt{5} \times 1)$
$\Rightarrow 30 = 2 \sqrt{5} \times 3 \sqrt{5}$

Now,
$x^{2} + 5 \sqrt{5} x + 30$

$= x^{2} + (3 \sqrt{5} + 2 \sqrt{5}) x + 30$ ... $[∵ 30 = 3 \sqrt{5} \times 2 \sqrt{5}, 5 \sqrt{5} = 3 \sqrt{5} + 2 \sqrt{5}]$

$= x^{2} + 3 \sqrt{5} x + 2 \sqrt{5} x + 30$

$= x (x + 3 \sqrt{5}) + 2 \sqrt{5} (x + 3 \sqrt{5})$

$= (x + 3 \sqrt{5}) (x + 2 \sqrt{5})$

The question number 14 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

14. $x^{2} - 24 x - 180$

Answer

We have
$x^{2} - 24 x - 180$

Here,
$180 = 2 \times 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 180 = (2 \times 3 \times 5) \times (2 \times 3 \times 1)$
$\Rightarrow 180 = 30 \times 6$

Now,
$x^{2} - 24 x - 180$

$= x^{2} + (- 30 + 6) x - 180$ ... $[∵ - 180 = (- 30) \times 6, - 24 = - 30 + 6]$

$= x^{2} - 30 x + 6 x - 180$

$= x (x - 30) + 6 (x - 30)$

$= (x - 30) (x + 6)$

The question number 15 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

15. $x^{2} - 32 x - 105$

Answer

We have
$x^{2} - 32 x - 105$

Here
$105 = 3 \times 5 \times 7 \times 1$
$\Rightarrow 105 = 3 \times (5 \times 7 \times 1)$
$\Rightarrow 105 = 3 \times 35$

Now,
$x^{2} - 32 x - 105$

$= x^{2} + (- 35 + 3) x - 105$ ... $[∵ - 105 = (- 35) \times 3, - 32 = - 35 + 3]$

$= x^{2} - 35 x + 3 x - 105$

$= x (x - 35) + 3 (x - 35)$

$= (x - 35) (x + 3)$

The question number 16 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

16. $x^{2} - 11 x - 80$

Answer

We have
$x^{2} - 11 x - 80$

Here,
$80 = 2 \times 2 \times 2 \times 2 \times 5 \times 1$
$\Rightarrow 80 = (2 \times 2 \times 2 \times 2) \times (5 \times 1)$
$\Rightarrow 80 = 16 \times 5$

Now,
$x^{2} - 11 x - 80$

$= x^{2} + (- 16 + 5) x - 80$ ... $[∵ - 80 = (- 16) \times 5, - 11 = - 16 + 5]$

$= x^{2} - 16 x + 5 x - 80$

$= x (x - 16) + 5 (x - 16)$

$= (x - 16) (x + 5)$

The question number 17 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

17. $6 - x - x^{2}$

Answer

We have
$6 - x - x^{2}$

Method 1:
$6 - x - x^{2}$
$= 6 - 3 x + 2 x - x^{2}$
$= 3 (2 - x) + x (2 - x)$
$= (3 + x) (2 - x)$

Method 2:
$6 - x - x^{2}$

$= - x^{2} - x + 6$

$= - (x^{2} + x - 6)$

$= - [x^{2} + (3 - 2) x - 6]$ ... $[∵ - 6 = 3 \times (- 2), 1 = 3 - 2]$

$= - [x^{2} + 3 x - 2 x - 6]$

$= - [x (x + 3) - 2 (x + 3)]$

$= - [(x - 2) (x + 3)]$

$= - (x - 2) (x + 3)$

$= (2 - x) (x + 3)$

$= (2 - x) (3 + x)$

The question number 18 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

18. $x^{2} - \sqrt{3} x - 6$

Answer

We have
$x^{2} - \sqrt{3} x - 6$

Here,
$6 = 2 \times 3 \times 1$
$\Rightarrow 6 = 2 \times \sqrt{3} \times \sqrt{3} \times 1$
$\Rightarrow 6 = 2 \sqrt{3} \times \sqrt{3}$

Now,
$x^{2} - \sqrt{3} x - 6$

$= x^{2} + (- 2 \sqrt{3} + \sqrt{3}) x - 6$ ... $[∵ - 6 = - 2 \sqrt{3} \times \sqrt{3}, \sqrt{3} = - 2 \sqrt{3} + \sqrt{3}]$

$= x^{2} - 2 \sqrt{3} x + \sqrt{3} x - 6$

$= x (x - 2 \sqrt{3}) + \sqrt{3} (x - 2 \sqrt{3})$

$= (x - 2 \sqrt{3}) (x + \sqrt{3})$

The question number 19 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

19. $40 + 3 x - x^{2}$

Answer

We have
$40 + 3 x - x^{2}$

Method 1:
$40 + 3 x - x^{2}$
$= 40 + (8 - 5) x - x^{2}$
$= 40 + 8 x - 5 x - x^{2}$
$= 8 (5 + x) - x (5 + x)$
$= (8 - x) (5 + x)$

Method 2:
$40 + 3 x - x^{2}$

$= - x^{2} + 3 x + 40$

$= - (x^{2} - 3 x - 40)$

Here,
$40 = 2 \times 2 \times 2 \times 5 \times 1$
$\Rightarrow 40 = (2 \times 2 \times 2) \times (5 \times 1)$
$\Rightarrow 40 = 8 \times 5$

Now,
$- (x^{2} - 3 x - 40)$

$= - [x^{2} + (- 8 + 5) x - 40]$ ... $[∵ - 40 = - 8 \times 5, - 3 = - 8 + 5]$

$= - [x^{2} - 8 x + 5 x - 40]$

$= - [x (x - 8) + 5 (x - 8)]$

$= - (x - 8) (x + 5)$

$= (8 - x) (5 + x)$

The question number 20 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

20. $x^{2} - 26 x + 133$

Answer

We have
$x^{2} - 26 x + 133$

Here,
$133 = 7 \times 19 \times 1$
$\Rightarrow 133 = 7 \times 19$

Now,
$x^{2} - 26 x + 133$

$= x^{2} + (- 19 - 7) x + 133$ ... $[∵ - 26 = - 19 - 7, 133 = (- 19) \times (- 7)]$

$= x^{2} - 19 x - 7 x + 133$

$= x (x - 19) - 7 (x - 19)$

$= (x - 19) (x - 7)$

The question number 21 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

21. $x^{2} - 2 \sqrt{3} x - 24$

Answer

We have
$x^{2} - 2 \sqrt{3} x - 24$

Here,
$24 = 2 \times 2 \times 2 \times 3 \times 1$
$\Rightarrow 24 = 2 \times 2 \times 2 \times \sqrt{3} \times \sqrt{3} \times 1$
$\Rightarrow 24 = (2 \times 2 \times \sqrt{3}) \times (2 \times \sqrt{3} \times 1)$
$\Rightarrow 24 = 4 \sqrt{3} \times 2 \sqrt{3}$

Now,
$x^{2} - 2 \sqrt{3} x - 24$

$= x^{2} + (- 4 \sqrt{3} + 2 \sqrt{3}) x - 24$ ... $[∵ - 2 \sqrt{3} = - 4 \sqrt{3} + 2 \sqrt{3}, - 24 = (- 4 \sqrt{3}) \times (2 \sqrt{3})]$

$= x^{2} - 4 \sqrt{3} x + 2 \sqrt{3} x - 24$

$= x (x - 4 \sqrt{3}) + 2 \sqrt{3} (x - 4 \sqrt{3})$

$= (x - 4 \sqrt{3}) (x + 2 \sqrt{3})$

The question number 22 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

22. $x^{2} - 3 \sqrt{5} x - 20$

Answer

We have
$x^{2} - 3 \sqrt{5} x - 20$

Here,
$20 = 2 \times 2 \times 5 \times 1$
$\Rightarrow 20 = 2 \times 2 \times \sqrt{5} \times \sqrt{5} \times 1$
$\Rightarrow 20 = (2 \times 2 \times \sqrt{5}) \times (\sqrt{5} \times 1)$
$\Rightarrow 20 = 4 \sqrt{5} \times \sqrt{5}$

Now,
$x^{2} - 3 \sqrt{5} x - 20$

$= x^{2} - 4 \sqrt{5} x + \sqrt{5} x - 20$ ... $[∵ - 2 \sqrt{5} = - 4 \sqrt{5} + \sqrt{5}, - 20 = (- 4 \sqrt{5}) \times (\sqrt{5})]$

$= x (x - 4 \sqrt{5}) + \sqrt{5} (x - 4 \sqrt{5})$

$= (x - 4 \sqrt{5}) (x + \sqrt{5})$

The question number 23 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

23. $x^{2} + \sqrt{2} x - 24$

Answer

We have
$x^{2} + \sqrt{2} x - 24$

Here,
$24 = 2 \times 2 \times 2 \times 3 \times 1$
$\Rightarrow 24 = 2 \times 2 \times \sqrt{2} \times \sqrt{2} \times 3 \times 1$
$\Rightarrow 24 = (2 \times 2 \times \sqrt{2}) \times (\sqrt{2} \times 3 \times 1)$
$\Rightarrow 24 = 4 \sqrt{2} \times 3 \sqrt{2}$

Now,
$x^{2} + \sqrt{2} x - 24$

$= x^{2} + 4 \sqrt{2} x - 3 \sqrt{2} x - 24$ ... $[∵ \sqrt{2} = 4 \sqrt{2} - 3 \sqrt{2}, - 24 = (4 \sqrt{2}) \times (- 3 \sqrt{2})]$

$= x (x + 4 \sqrt{2}) - 3 \sqrt{2} (x + 4 \sqrt{2})$

$= (x + 4 \sqrt{2}) (x - 3 \sqrt{2})$

The question number 24 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

24. $x^{2} - 2 \sqrt{2} x - 30$

Answer

We have
$x^{2} - 2 \sqrt{2} x - 30$

Here,
$30 = 2 \times 3 \times 5 \times 1$
$\Rightarrow 30 = \sqrt{2} \times \sqrt{2} \times 3 \times 5 \times 1$
$\Rightarrow 30 = 5 \sqrt{2} \times 3 \sqrt{2}$

Now
$x^{2} - 2 \sqrt{2} x - 30$

$= x^{2} + (- 5 \sqrt{2} + 3 \sqrt{2}) x - 30$ ... $[∵ - 2 \sqrt{2} = - 5 \sqrt{2} + 3 \sqrt{2}, - 30 = (- 5 \sqrt{2}) \times (3 \sqrt{2})]$

$= x^{2} - 5 \sqrt{2} x + 3 \sqrt{2} x - 30$

$= x (x - 5 \sqrt{2}) + 3 \sqrt{2} (x - 5 \sqrt{2})$

$= (x + 3 \sqrt{2}) (x - 5 \sqrt{2})$

The question number 25 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

25. $x^{2} - x - 156$

Answer

We have
$x^{2} - x - 156$

Here,
$156 = 2 \times 2 \times 3 \times 13 \times 1$
$\Rightarrow 156 = (2 \times 2 \times 3) \times (13 \times 1)$
$\Rightarrow 156 = 12 \times 13$

Now,
$x^{2} - x - 156$

$= x^{2} + (- 13 + 12) x - 156$ ... $[∵ - 1 = - 13 + 12, - 156 = (- 13) \times 12]$

$= x^{2} - 13 x + 12 x - 156$

$= x (x - 13) + 12 (x - 13)$

$= (x - 13) (x + 12)$

The question number 26 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

26. $x^{2} - 32 x - 105$

Answer

We have
$x^{2} - 32 x - 105$

Here
$105 = 3 \times 5 \times 7 \times 1$
$\Rightarrow 105 = 3 \times (5 \times 7 \times 1)$
$\Rightarrow 105 = 3 \times 35$

Now,
$x^{2} - 32 x - 105$

$= x^{2} + (- 35 + 3) x - 105$ ... $[∵ - 32 = - 35 + 3, - 105 = (- 35) \times 3]$

$= x^{2} - 35 x + 3 x - 105$

$= x (x - 35) + 3 (x - 35)$

$= (x - 35) (x + 3)$

The question number 27 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

27. $9 x^{2} + 18 x + 8$

Answer

We have
$9 x^{2} + 18 x + 8$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 9 \times 8 = 72$
And,
$72 = 2 \times 2 \times 2 \times 3 \times 3 \times 1$
$\Rightarrow 72 = (2 \times 2 \times 3) \times (2 \times 3 \times 1)$
$\Rightarrow 72 = 12 \times 6$

Now,
$9 x^{2} + 18 x + 8$

$= 9 x^{2} + (12 + 6) x + 8$ ... $[∵ 18 = 12 + 6, 72 = 12 \times 6]$

$= 9 x^{2} + 12 x + 6 x + 8$

$= 3 x (3 x + 4) + 2 (3 x + 4)$

$= (3 x + 4) (3 x + 2)$

The question number 28 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

28. $6 x^{2} + 17 x + 12$

Answer

We have
$6 x^{2} + 17 x + 12$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 6 \times 12 = 72$
And,
$72 = 2 \times 2 \times 2 \times 3 \times 3 \times 1$
$\Rightarrow 72 = (2 \times 2 \times 2) \times (3 \times 3 \times 1)$
$\Rightarrow 72 = 8 \times 9$

Now,
$6 x^{2} + 17 x + 12$

$= 6 x^{2} + (9 + 8) x + 12$ ... $[∵ 17 = 9 + 8, 72 = 9 \times 8]$

$= 6 x^{2} + 9 x + 8 x + 12$

$= 3 x (2 x + 3) + 4 (2 x + 3)$

$= (3 x + 4) (2 x + 3)$

The question number 29 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

29. $18 x^{2} + 3 x - 10$

Answer

We have
$18 x^{2} + 3 x - 10$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 18 \times (- 10) = - 180$
And,
$180 = 2 \times 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 180 = (2 \times 2 \times 3) \times (3 \times 5 \times 1)$
$\Rightarrow 180 = 12 \times 15$

Now,
$18 x^{2} + 3 x - 10$

$= 18 x^{2} + (15 - 12) x - 10$ ... $[∵ 3 = 15 - 12, - 180 = 15 \times (- 12)]$

$= 18 x^{2} + 15 x - 12 x - 10$

$= 3 x (6 x + 5) - 2 (6 x + 5)$

$= (6 x + 5) (3 x - 2)$

The question number 30 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

30. $2 x^{2} + 11 x - 21$

Answer

We have
$2 x^{2} + 11 x - 21$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \times (- 21) = - 42$
And,
$42 = 2 \times 3 \times 7 \times 1$
$\Rightarrow 42 = (2 \times 7) \times 3$
$\Rightarrow 42 = 14 \times 3$

Now,
$2 x^{2} + 11 x - 21$

$= 2 x^{2} + (14 - 3) x - 21$ ... $[∵ 11 = 14 - 3, - 42 = 14 \times (- 3)]$

$= 2 x^{2} + 14 x - 3 x - 21$

$= 2 x (x + 7) - 3 (x + 7)$

$= (2 x - 3) (x + 7)$

The question number 31 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

31. $15 x^{2} + 2 x - 8$

Answer

We have
$15 x^{2} + 2 x - 8$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 15 \times (- 8) = - 120$
And,
$120 = 2 \times 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 120 = (2 \times 2 \times 3) \times (2 \times 5 \times 1)$
$\Rightarrow 120 = 12 \times 10$

Now,
$15 x^{2} + 2 x - 8$

$= 15 x^{2} + (12 - 10) x - 8$ ... $[∵ 2 = 12 - 10, - 120 = 12 \times (- 10)]$

$= 15 x^{2} + 12 x - 10 x - 8$

$= 3 x (5 x + 4) - 2 (5 x + 4)$

$= (3 x - 2) (5 x + 4)$

The question number 32 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

32. $21 x^{2} + 5 x - 6$

Answer

We have
$21 x^{2} + 5 x - 6$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 21 \times (- 6) = - 126$
And,
$126 = 2 \times 3 \times 3 \times 7 \times 1$
$\Rightarrow 126 = (2 \times 7) \times (3 \times 3 \times 1)$
$\Rightarrow 126 = 14 \times 9$

Now,
$21 x^{2} + 5 x - 6$

$= 21 x^{2} + (14 - 9) x - 6$ ... $[∵ 5 = 14 - 9, - 126 = 14 \times (- 9)]$

$= 21 x^{2} + 14 x - 9 x - 6$

$= 7 x (3 x + 2) - 3 (3 x + 2)$

$= (3 x + 2) (7 x - 3)$

The question number 33 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

33. $24 x^{2} - 41 x + 12$

Answer

We have
$24 x^{2} - 41 x + 12$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 24 \times 12 = < 288$
And,
$288 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 1$
$\Rightarrow 288 = (2 \times 2 \times 2 \times 2 \times 2) \times (3 \times 3 \times 1)$
$\Rightarrow 288 = 32 \times 9$

Now,
$24 x^{2} - 41 x + 12$

$= 24 x^{2} + (- 32 - 9) x + 12$ ... $[∵ - 41 = - 32 - 9, 288 = (- 12) \times (- 9)]$

$= 24 x^{2} - 32 x - 9 x + 12$

$= 8 x (3 x - 4) - 3 (3 x - 4)$

$= (3 x - 4) (8 x - 3)$

The question number 34 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

34. $3 x^{2} - 14 x + 8$

Answer

We have
$3 x^{2} - 14 x + 8$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 3 \times 8 = < 24$
And,
$24 = 2 \times 2 \times 2 \times 3 \times 1$
$\Rightarrow 24 = (2 \times 2 \times 3) \times (2 \times 1)$
$\Rightarrow 24 = 12 \times 2$

Now,
$3 x^{2} - 14 x + 8$

$= 3 x^{2} + (- 12 - 2) x + 8$ ... $[∵ - 14 = - 12 - 2, 24 = (- 12) \times (- 2)]$

$= 3 x^{2} - 12 x - 2 x + 8$

$= 3 x (x - 4) - 2 (x - 4)$

$= (x - 4) (3 x - 2)$

The question number 35 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

35. $2 x^{2} + 3 x - 90$

Answer

We have
$2 x^{2} + 3 x - 90$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \times (- 90) = - 180$
And,
$180 = 2 \times 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 180 = (2 \times 2 \times 3) \times (3 \times 5 \times 1)$
$\Rightarrow 180 = 12 \times 15$

Now,
$2 x^{2} + 3 x - 90$

$= 2 x^{2} + (15 - 12) x - 90$ ... $[∵ 3 = 15 - 12, 90 = 15 \times 12]$

$= 2 x^{2} + 15 x - 12 x - 90$

$= x (2 x + 15) - 6 (2 x + 15)$

$= (2 x + 15) (x - 6)$

The question number 36 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

36. $\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$

Answer

We have
$\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= \sqrt{5} \times (- 3 \sqrt{5})$ $= - 15$
And,
$15 = 3 \times 5 \times 1$
$\Rightarrow 15 = 3 \times (5 \times 1)$
$\Rightarrow 15 = 3 \times 5$

Now,
$\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$

$= \sqrt{5} x^{2} + (5 - 3) x - 3 \sqrt{5}$ ... $[∵ 2 = 5 - 3, - 15 = 5 \times (- 3)]$

$= \sqrt{5} x^{2} + 5 x - 3 x - 3 \sqrt{5}$

$= \sqrt{5} x (x + \sqrt{5}) - 3 (x + \sqrt{5})$

$= (x + \sqrt{5}) (\sqrt{5} x - 3)$

The question number 37 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

37. $2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$

Answer

We have
$2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \sqrt{3} \times (- 5 \sqrt{3}) = - 30$
And,
$30 = 2 \times 3 \times 5 \times 1$
$\Rightarrow 30 = (2 \times 3) \times (5 \times 1)$
$\Rightarrow 30 = 6 \times 5$

Now,
$2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$

$= 2 \sqrt{3} x^{2} + (6 - 5) x - 5 \sqrt{3}$ ... $[∵ 1 = 6 - 5, - 30 = 6 \times (- 5)]$

$= 2 \sqrt{3} x^{2} + 6 x - 5 x - 5 \sqrt{3}$

$= 2 \sqrt{3} x (x + \sqrt{3}) - 5 (x + \sqrt{3})$

$= (x + \sqrt{3}) (2 \sqrt{3} x - 5)$

The question number 38 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

38. $7 x^{2} + 2 \sqrt{14} x + 2$

Answer

We have
$7 x^{2} + 2 \sqrt{14} x + 2$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 7 \times 2 = < 14$
And,
$14 = \sqrt{14} \times \sqrt{14}$

Now,
$7 x^{2} + 2 \sqrt{14} x + 2$

$= 7 x^{2} + (\sqrt{14} + \sqrt{14}) x + 2$ ... $[∵ 2 \sqrt{14} = \sqrt{14} + \sqrt{14}, 14 = \sqrt{14} \times \sqrt{14}]$

$= 7 x^{2} + \sqrt{14} x + \sqrt{14} x + 2$

$= \sqrt{7} \times \sqrt{7} x^{2} + \sqrt{7} \times \sqrt{2} x + \sqrt{7} \times \sqrt{2} x + \sqrt{2} \times \sqrt{2}$

$= \sqrt{7} x (\sqrt{7} x + \sqrt{2}) + \sqrt{2} (\sqrt{7} x + \sqrt{2})$

$= (\sqrt{7} x + \sqrt{2}) (\sqrt{7} x + \sqrt{2})$

The question number 39 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

39. $6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

Answer

We have
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 6 \sqrt{3} \times 5 \sqrt{3} = 90$
And,
$90 = 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 90 = 2 \times (3 \times 3 \times 5 \times 1)$
$\Rightarrow 90 = 2 \times 45$

Now,
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

$= 6 \sqrt{3} x^{2} - (45 + 2) x + 5 \sqrt{3}$ ... $[∵ - 47 = - 45 - 2, 90 = (- 45) \times (- 2)]$

$= 6 \sqrt{3} x^{2} - 45 x - 2 x + 5 \sqrt{3}$

$= 3 \sqrt{3} x (2 x - 5 \sqrt{3}) - 1 (2 x - 5 \sqrt{3})$

$= (2 x - 5 \sqrt{3}) (3 \sqrt{3} x - 1)$

The question number 40 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

40. $5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Answer

We have
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 5 \sqrt{5} \times 3 \sqrt{5} = 75$
And,
$75 = 3 \times 5 \times 5 \times 1$
$\Rightarrow 75 = (3 \times 5) \times (5 \times 1)$
$\Rightarrow 75 = 15 \times 5$

Now,
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

$= 5 \sqrt{5} x^{2} + (15 + 5) x + 3 \sqrt{5}$ ... $[∵ 20 = 15 + 5, 75 = 15 \times 5]$

$= 5 \sqrt{5} x^{2} + 15 x + 5 x + 3 \sqrt{5}$

$= 5 x (\sqrt{5} x + 3) + \sqrt{5} (\sqrt{5} x + 3)$

$= (\sqrt{5} x + 3) (5 x + \sqrt{5})$

The question number 41 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

41. $\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

Answer

We have
$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= \sqrt{3} \times 8 \sqrt{3} = 24$
And,
$24 = 2 \times 2 \times 2 \times 3 \times 1$
$\Rightarrow 24 = (2 \times 2) \times (2 \times 3 \times 1)$
$\Rightarrow 24 = 4 \times 6$

Now,
$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

$= \sqrt{3} x^{2} + (6 + 4) x + 8 \sqrt{3}$ ... $[∵ 10 = 6 + 4, 24 = 6 \times 4]$

$= \sqrt{3} x^{2} + 6 x + 4 x + 8 \sqrt{3}$

$= \sqrt{3} x (x + 2 \sqrt{3}) + 4 (x + 2 \sqrt{3})$

$= (x + 2 \sqrt{3}) (\sqrt{3} x + 4)$

The question number 42 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

42. $\sqrt{2} x^{2} + 3 x + \sqrt{2}$

Answer

We have,
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= \sqrt{2} \times \sqrt{2} = 2$
And,
$2 = 2 \times 1$

Now,
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$

$= \sqrt{2} x^{2} + (2 + 1) x + \sqrt{2}$ ... $[∵ 3 = 2 + 1, 24 = 2 \times 1]$

$= \sqrt{2} x^{2} + 2 x + x + \sqrt{2}$

$= \sqrt{2} x (x + \sqrt{2}) + 1 (x + \sqrt{2})$

$= (x + \sqrt{2}) (\sqrt{2} x + 1)$

The question number 43 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

43. $2 x^{2} + 3 \sqrt{3} x + 3$

Answer

We have
$2 x^{2} + 3 \sqrt{3} x + 3$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \times 3 = 6$
And,
$6 = 2 \times 3 \times 1$
$\Rightarrow 6 = 2 \times \sqrt{3} \times \sqrt{3} \times 1$
$\Rightarrow 6 = (2 \times \sqrt{3}) \times (\sqrt{3} \times 1)$
$\Rightarrow 6 = 2 \sqrt{3} \times \sqrt{3}$

Now,
$2 x^{2} + 3 \sqrt{3} x + 3$

$= 2 x^{2} + (2 \sqrt{3} + \sqrt{3}) x + 3$ ... $[∵ 3 \sqrt{3} = 2 \sqrt{3} + \sqrt{3}, 6 = 2 \sqrt{3} \times \sqrt{3}]$

$= 2 x^{2} + 2 \sqrt{3} x + \sqrt{3} x + 3$

$= 2 x (x + \sqrt{3}) + \sqrt{3} (x + \sqrt{3})$

$= (x + \sqrt{3}) (2 x + \sqrt{3})$

The question number 44 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

44. $15 x^{2} - x - 28$

Answer

We have
$15 x^{2} - x - 28$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 15 \times (- 28) = - 420$
And,
$420 = 2 \times 2 \times 3 \times 5 \times 7 \times 1$
$\Rightarrow 420 = (2 \times 2 \times 5) \times (3 \times 7 \times 1)$
$\Rightarrow 420 = 20 \times 21$

Now,
$15 x^{2} - x - 28$

$= 15 x^{2} + (- 21 + 20) x - 28$ ... $[∵ - 1 = - 21 + 20, - 420 = - 21 \times 20]$

$= 15 x^{2} - 21 x + 20 x - 28$

$= 3 x (5 x - 7) + 4 (5 x - 7)$

$= (5 x - 7) (3 x + 4)$

The question number 45 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

45. $6 x^{2} - 5 x - 21$

Answer

We have
$6 x^{2} - 5 x - 21$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 6 \times (- 21) = - 126$
And,
$126 = 2 \times 3 \times 3 \times 7 \times 1$
$\Rightarrow 126 = (2 \times 7) \times (3 \times 3 \times 1)$
$\Rightarrow 126 = 14 \times 9$

Now,
$6 x^{2} - 5 x - 21$

$= 6 x^{2} + (- 14 + 9) x - 21$ ... $[∵ - 5 = - 14 + 9, - 126 = - 14 \times 9]$

$= 6 x^{2} - 14 x + 9 x - 21$

$= 2 x (3 x - 7) + 3 (3 x - 7)$

$= (3 x - 7) (2 x + 3)$

The question number 46 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

46. $2 x^{2} - 7 x - 15$

Answer

We have
$2 x^{2} - 7 x - 15$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \times (- 15) = - 30$
And,
$30 = 2 \times 5 \times 3 \times 1$
$\Rightarrow 30 = (2 \times 5) \times (3 \times 1)$
$\Rightarrow 30 = 10 \times 3$

Now,
$2 x^{2} - 7 x - 15$

$= 2 x^{2} + (- 10 + 3) x - 15$ ... $[∵ - 7 = - 10 + 3, - 30 = - 10 \times 3]$

$= 2 x^{2} - 10 x + 3 x - 15$

$= 2 x (x - 5) + 3 (x - 5)$

$= (x - 5) (2 x + 3)$

The question number 47 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

47. $5 x^{2} - 16 x - 21$

Answer

We have
$5 x^{2} - 16 x - 21$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 5 \times (- 21) = - 105$
And,
$105 = 3 \times 5 \times 7 \times 1$
$\Rightarrow 105 = (3 \times 7) \times (5 \times 1)$
$\Rightarrow 105 = 21 \times 5$

Now,
$5 x^{2} - 16 x - 21$

$= 5 x^{2} + (- 21 + 5) x - 21$ ... $[∵ - 16 = - 21 + 5, - 105 = - 21 \times 5]$

$= 5 x^{2} - 21 x + 5 x - 21$

$= x (5 x - 21) + 1 (5 x - 21)$

$= (5 x - 21) (x + 1)$

The question number 48 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

48. $6 x^{2} - 11 x - 35$

Answer

We have
$6 x^{2} - 11 x - 35$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 6 \times (- 35) = - 210$
And,
$180 = 2 \times 2 \times 3 \times 3 \times 5 \times 1$
$\Rightarrow 210 = (2 \times 5) \times (3 \times 7 \times 1)$
$\Rightarrow 210 = 10 \times 21$

Now,
$6 x^{2} - 11 x - 35$

$= 6 x^{2} + (- 21 + 10) x - 35$ ... $[∵ - 11 = - 21 + 10, - 210 = - 21 \times 10]$

$= 6 x^{2} - 21 x + 10 x - 35$

$= 3 x (2 x - 7) + 5 (2 x - 7)$

$= (2 x - 7) (3 x + 5)$

The question number 49 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

49. $9 x^{2} - 3 x - 20$

Answer

We have
$9 x^{2} - 3 x - 20$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 9 \times (- 20) = - 180$
And,
$180 = 2 \times 3 \times 5 \times 7 \times 1$
$\Rightarrow 180 = (2 \times 2 \times 3) \times (3 \times 5 \times 1)$
$\Rightarrow 180 = 12 \times 15$

Now,
$9 x^{2} - 3 x - 20$

$= 9 x^{2} + (- 15 + 12) x - 20$ ... $[∵ - 3 = - 15 + 12, - 180 = - 15 \times 12]$

$= 9 x^{2} - 15 x + 12 x - 20$

$= 3 x (3 x - 5) + 4 (3 x - 5)$

$= (3 x - 5) (3 x + 4)$

The question number 50 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

50. $10 x^{2} - 9 x - 7$

Answer

We have
$10 x^{2} - 9 x - 7$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 10 \times (- 7) = - 70$
And,
$70 = 2 \times 5 \times 7 \times 1$
$\Rightarrow 70 = (2 \times 7) \times (5 \times 1)$
$\Rightarrow 70 = 14 \times 5$

Now,
$10 x^{2} - 9 x - 7$

$= 10 x^{2} + (- 14 + 5) x - 7$ ... $[∵ - 9 = - 14 + 5, - 70 = - 14 \times 5]$

$= 10 x^{2} - 14 x + 5 x - 7$

$= 2 x (5 x - 7) + 1 (5 x - 7)$

$= (5 x - 7) (2 x + 1)$

The question number 51 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

51. $x^{2} - 2 x + \frac{7}{16}$

Answer

We have
$x^{2} - 2 x + \frac{7}{16}$

$= \frac{1}{16} [16 x^{2} - 32 x + 7]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 16 \times 7 = 112$
And,
$112 = 2 \times 2 \times 2 \times 2 \times 7 \times 1$
$\Rightarrow 112 = (2 \times 2) \times (2 \times 2 \times 7 \times 1)$
$\Rightarrow 112 = 4 \times 28$

Now,
$\frac{1}{16} [16 x^{2} - 32 x + 7]$

$= \frac{1}{16} [16 x^{2} + (- 28 - 4) x + 7]$ ... $[∵ - 32 = - 28 - 4, 112 = (- 28) \times (- 4)]$

$= \frac{1}{16} [16 x^{2} - 28 x - 4 x + 7]$

$= \frac{1}{16} [4 x (4 x - 7) - 1 (4 x - 7)]$

$= \frac{1}{16} (4 x - 7) (4 x - 1)$

$= (\frac{4 x}{16} - \frac{7}{16}) (4 x - 1)$

$= (\frac{x}{4} - \frac{7}{16}) (4 x - 1)$

The question number 52 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

52. $\frac{1}{3} x^{2} - 2 x - 9$

Answer

We have
$\frac{1}{3} x^{2} - 2 x - 9$

$= \frac{1}{3} [x^{2} - 6 x - 27]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 1 \times (- 27) = - 27$
And,
$27 = 3 \times 3 \times 3 \times 1$
$\Rightarrow 27 = (3 \times 3) \times (3 \times 1)$
$\Rightarrow 27 = 9 \times 3$

Now,
$\frac{1}{3} [x^{2} - 6 x - 27]$

$= \frac{1}{3} [x^{2} + (- 9 + 3) x - 27]$ ... $[∵ - 6 = - 9 + 3, - 27 = (- 9) \times 3]$

$= \frac{1}{3} [x^{2} - 9 x + 3 x - 27]$

$= \frac{1}{3} [x (x - 9) + 3 (x - 9)]$

$= \frac{1}{3} (x + 3) (x - 9)$

$= (\frac{x}{3} + \frac{3}{3}) (x - 9)$

$= (\frac{x}{3} + 1) (x - 9)$

The question number 53 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

53. $x^{2} + \frac{12}{35} x + \frac{1}{35}$

Answer

We have
$x^{2} + \frac{12}{35} x + \frac{1}{35}$

$= \frac{1}{35} [35 x^{2} + 12 x + 1]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 35 \times 1 = 35$
And,
$35 = 5 \times 7 \times 1$
$\Rightarrow 35 = 5 \times 7$

Now,
$\frac{1}{35} [35 x^{2} + 12 x + 1]$

$= \frac{1}{35} [35 x^{2} + 7 x + 5 x + 1]$

$= \frac{1}{35} [7 x (5 x + 1) + 1 (5 x + 1)]$

$= \frac{1}{35} (5 x + 1) (7 x + 1)$

$= (5 x + 1) (\frac{7 x}{35} + \frac{1}{35})$

$= (5 x + 1) (\frac{x}{5} + \frac{1}{35})$

The question number 54 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

54. $21 x^{2} - 2 x + \frac{1}{21}$

Answer

We have
$21 x^{2} - 2 x + \frac{1}{21}$

$= \frac{1}{21} [441 x^{2} - 42 x + 1]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 441 \times 1 = 441$
And,
$441 = 3 \times 3 \times 7 \times 7 \times 1$
$\Rightarrow 441 = (3 \times 7) \times (3 \times 7 \times 1)$
$\Rightarrow 441 = 21 \times 21$

Now,
$\frac{1}{21} [441 x^{2} - 42 x + 1]$

$= \frac{1}{21} [441 x^{2} + (- 21 - 21) x + 1]$ ... $[∵ - 42 = - 21 - 21, 441 = (- 21) \times (- 21)]$

$= \frac{1}{21} [441 x^{2} - 21 x - 21 x + 1]$

$= \frac{1}{21} [21 x (21 x - 1) - 1 (21 x - 1)]$

$= \frac{1}{21} (21 x - 1) (21 x - 1)$

$= (\frac{21 x}{21} - \frac{1}{21}) (21 x - 1)$

$= (x - \frac{1}{21}) (21 x - 1)$

The question number 55 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

55. $\frac{3}{2} x^{2} + 16 x + 10$

Answer

We have
$\frac{3}{2} x^{2} + 16 x + 10$

$= \frac{1}{2} [3 x^{2} + 32 x + 20]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 3 \times 20 = 60$
And,
$60 = 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 60 = 2 \times (2 \times 3 \times 5 \times 1)$
$\Rightarrow 60 = 2 \times 30$

Now,
$\frac{1}{2} [3 x^{2} + 32 x + 20]$

$= \frac{1}{2} [3 x^{2} + (30 + 2) x + 20]$ ... $[∵ 32 = 30 + 2, 60 = 30 \times 2]$

$= \frac{1}{2} [3 x^{2} + 30 x + 2 x + 20]$

$= \frac{1}{2} [3 x (x + 10) + 2 (x + 10)]$

$= \frac{1}{2} (x + 10) (3 x + 2)$

$= (\frac{x}{2} + \frac{10}{2}) (3 x + 2)$

$= (\frac{x}{2} + 5) (3 x + 2)$

The question number 56 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

56. $\frac{2}{3} x^{2} - \frac{17}{3} x - 28$

Answer

We have
$\frac{2}{3} x^{2} - \frac{17}{3} x - 28$

$= \frac{1}{3} [2 x^{2} - 17 x - 84]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 2 \times (- 84) = - 168$
And,
$168 = 2 \times 2 \times 2 \times 3 \times 7 \times 1$
$\Rightarrow 168 = (2 \times 2 \times 2 \times 3) \times (7 \times 1)$
$\Rightarrow 168 = 24 \times 7$

Now,
$\frac{1}{3} [2 x^{2} - 17 x - 84]$

$= \frac{1}{3} [2 x^{2} + (- 24 + 7) x - 84]$ ... $[∵ - 17 = - 24 + 7, - 168 = - 24 \times 7]$

$= \frac{1}{3} [2 x^{2} - 24 x + 7 x - 84]$

$= \frac{1}{3} [2 x (x - 12) + 7 (x - 12)]$

$= \frac{1}{3} (x - 12) (2 x + 7)$

$= (\frac{x}{3} - \frac{12}{3}) (2 x + 7)$

$= (\frac{x}{3} - 4) (2 x + 7)$

The question number 57 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

57. $\frac{3}{5} x^{2} - \frac{19}{5} x + 4$

Answer

We have
$\frac{3}{5} x^{2} - \frac{19}{5} x + 4$

$= \frac{1}{5} [3 x^{2} - 19 x + 20]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 3 \times 20 = 60$
And,
$60 = 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 60 = (2 \times 2) \times (3 \times 5 \times 1)$
$\Rightarrow 60 = 4 \times 15$

$\frac{1}{5} [3 x^{2} - 19 x + 20]$

$= \frac{1}{5} [3 x^{2} + (- 15 - 4) x + 20]$ ... $[∵ - 17 = - 15 - 4, 60 = (- 15) \times (- 4)]$

$= \frac{1}{5} [3 x^{2} - 15 x - 4 x + 20]$

$= \frac{1}{5} [3 x (x - 5) - 4 (x - 5)]$

$= \frac{1}{5} (x - 5) (3 x - 4)$

$= (\frac{x}{5} - \frac{5}{5}) (3 x - 4)$

$= (\frac{x}{5} - 1) (3 x - 4)$

The question number 58 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

58. $2 x^{2} - x + \frac{1}{8}$

Answer

We have
$2 x^{2} - x + \frac{1}{8}$

$= \frac{1}{8} [16 x^{2} - 8 x + 1]$

Here,
∵ the coefficient of $x^{2}$ × constant term
$= 16 \times 1 = 16$
And,
$16 = 2 \times 2 \times 2 \times 2 \times 1$
$\Rightarrow 16 = (2 \times 2) \times (2 \times 2 \times 1)$
$\Rightarrow 16 = 4 \times 4$

Now,
$\frac{1}{8} [16 x^{2} - 8 x + 1]$

$= \frac{1}{8} [16 x^{2} + (- 4 - 4) x + 1]$ ... $[∵ - 8 = - 4 - 4, 16 = (- 4) \times (- 4)]$

$= \frac{1}{8} [16 x^{2} - 4 x - 4 x + 1]$

$= \frac{1}{8} [4 x (4 x - 1) - 1 (4 x - 1)]$

$= \frac{1}{8} (4 x - 1) (4 x - 1)$

$= (\frac{4 x}{8} - \frac{1}{8}) (4 x - 1)$

$= (\frac{x}{2} - \frac{1}{8}) (4 x - 1)$

The question number 59 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

59. $2 {(x + y)}^{2} - 9 (x + y) - 5$

Answer

We have
$2 {(x + y)}^{2} - 9 (x + y) - 5$

Let $x + y = v$ , the given polynomial becomes
$2 v^{2} - 9 v - 5$

Here,
∵ the coefficient of $v^{2}$ × constant term
$= 2 \times (- 5) = - 10$
And,
$10 = 2 \times 5 \times 1$
$\Rightarrow 10 = (2 \times 5) \times (1)$
$\Rightarrow 10 = 10 \times 1$

Now,
$2 v^{2} - 9 v - 5$

$= 2 v^{2} + (- 10 + 1) v - 5$ ... $[∵ - 9 = - 10 + 1, 10 = (- 10) \times 1]$

$= 2 v^{2} - 10 x + v - 5$

$= 2 v (v - 5) + 1 (v - 5)$

$= (v - 5) (2 v + 1)$

Putting $v = x + y$ , we get.

$= (x + y - 5) (2 x + 2 y + 1)$

The question number 60 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

60. $9 {(2 a - b)}^{2} - 4 (2 a - b) - 13$

Answer

We have
$9 {(2 a - b)}^{2} - 4 (2 a - b) - 13$

Let $2 a - b = y$ , the given polynomial becomes
$9 y^{2} - 4 y - 13$

Here,
∵ the coefficient of $y^{2}$ × constant term
$= 9 \times (- 13) = - 117$
And,
$117 = 3 \times 3 \times 13 \times 1$
$\Rightarrow 117 = (3 \times 3) \times (13 \times 1)$
$\Rightarrow 117 = 9 \times 13$

Now,
$9 y^{2} - 4 y - 13$

$= 9 y^{2} + (- 13 + 9) y - 13$ ... $[∵ - 4 = - 13 + 9, 117 = (- 13) \times 9]$

$= 9 y^{2} - 13 y + 9 y - 13$

$= y (9 y - 13) + 1 (9 y - 13)$

$= (9 y - 13) (y + 1)$

$= [9 (2 a - b) - 13] [(2 a - b) + 1]$ ... $[∵ y = (2 a - b)]$

$= (18 a - 9 b - 13) (2 a - b + 1)$

The question number 61 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

61. $7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$

Answer

We have
$7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$

Let $x - 2 y = z$ , the given polynomial becomes
$7 z^{2} - 25 z + 12$

Here,
∵ the coefficient of $z^{2}$ × constant term
$= 7 \times 12 = 84$
And,
$84 = 2 \times 2 \times 3 \times 7 \times 1$
$\Rightarrow 84 = (2 \times 2) \times (3 \times 7 \times 1)$
$\Rightarrow 84 = 4 \times 21$

Now,
$7 z^{2} - 25 z + 12$

$= 7 z^{2} + (- 21 - 4) z + 12$ ... $[∵ - 25 = - 21 - 4, 84 = (- 21) \times (- 4)]$

$= 7 z^{2} - 21 z - 4 z + 12$

$= 7 z (z - 3) - 4 (z - 3)$

$= (z - 3) (7 z - 4)$

$= [(x - 2 y) - 3] [7 (x - 2 y) - 4]$ ... $[∵ z = (x - 2 y)]$

$= (x - 2 y - 3) (7 x - 14 y - 4)$

The question number 62 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

62. $10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$

Answer

We have
$10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$

Let $(3 x + \frac{1}{x}) = y$ , the given polynomial becomes
$10 y^{2} - y - 3$

Here,
∵ the coefficient of $y^{2}$ × constant term
$= 10 \times (- 3) = - 30$
And,
$30 = 2 \times 3 \times 5 \times 1$
$\Rightarrow 30 = (2 \times 3) \times (5 \times 1)$
$\Rightarrow 30 = 6 \times 5$

Now,
$10 y^{2} - y - 3$

$= 10 y^{2} + (- 6 + 5) y - 3$ ... $[∵ - 1 = - 6 + 5, - 30 = (- 6) \times 5]$

$= 10 y^{2} - 6 y + 5 y - 3$

$= 2 y (5 y - 3) + 1 (5 y - 3)$

$= (2 y + 1) (5 y - 3)$

$= [2 (3 x + \frac{1}{x}) + 1] [5 (3 x + \frac{1}{x}) - 3]$ ... $[∵ y = (3 x + \frac{1}{x})]$

$= (6 x + \frac{2}{x} + 1) (15 x + \frac{5}{x} - 3)$

The question number 63 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

63. $6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$

Answer

We have
$6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$

Let $(2 x - \frac{3}{x}) = y$ , the given polynomial becomes
$6 y^{2} + 7 y - 20$

Here,
∵ the coefficient of $y^{2}$ × constant term
$= 6 \times (- 20) = - 120$
And,
$120 = 2 \times 2 \times 2 \times 3 \times 5 \times 1$
$\Rightarrow 120 = (2 \times 2 \times 2) \times (3 \times 5 \times 1)$
$\Rightarrow 120 = 8 \times 15$

Now,
$6 y^{2} + 7 y - 20$

$= 6 y^{2} + 15 y - 8 y - 20$ ... $[∵ 7 = 15 - 8, - 120 = 15 \times (- 8)]$

$= 3 y (2 y + 5) - 4 (2 y + 5)$

$= (2 y + 5) (3 y - 4)$

$= [2 (2 x - \frac{3}{x}) + 5] [3 (2 x - \frac{3}{x}) - 4]$ ... $[∵ y = (2 x - \frac{3}{x})]$

$= (4 x - \frac{6}{x} + 5) (6 x - \frac{9}{x} - 4)$

The question number 64 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

64. ${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$

Answer

We have
${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$

Let $(a + 2 b) = y$ , the given polynomial becomes
$y^{2} + 101 y + 100$

Here,
∵ the coefficient of $y^{2}$ × constant term
$= 1 \times 100 = 100$
And,
$100 = 2 \times 2 \times 5 \times 5 \times 1$
$\Rightarrow 100 = (2 \times 2 \times 5 \times 5) \times 1$
$\Rightarrow 100 = 100 \times 1$

Now,
$y^{2} + 101 y + 100$

$= y^{2} + (100 + 1) y + 100$ ... $[∵ 101 = 100 + 1, 100 = 100 \times 1]$

$= y^{2} + 100 y + y + 100$

$= y (y + 100) + 1 (y + 100)$

$= (y + 1) (y + 100)$

$= [(a + 2 b) + 1] [(a + 2 b) + 100]$ ... $[∵ y = (a + 2 b)]$

$= (a + 2 b + 1) (a + 2 b + 100)$

The question number 65 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

65. $4 x^{4} + 7 x^{2} - 2$

Answer

We have
$4 x^{4} + 7 x^{2} - 2$

$= 4 {(x^{2})}^{2} + 7 x^{2} - 2$

Let $x^{2} = y$ .

The given algebraic expression becomes
$4 y^{2} + 7 y - 2$

Here,
Coefficient of $y^{2}$ × Constant term
$= 4 \times (- 2) = - 8$

And,
$8 = 2 \times 2 \times 2 \times 1$
$\Rightarrow 8 = (2 \times 2 \times 2) \times 1$
$\Rightarrow 8 = 8 \times 1$

Now,
$4 y^{2} + 7 y - 2$

$= 4 y^{2} + (8 - 1) y - 2$ ... $[∵ 7 = 8 - 1, - 8 = 8 \times (- 1)]$

$= 4 y^{2} + 8 y - 1 y - 2$

$= 4 y (y + 2) - 1 (y + 2)$

$= (y + 2) (4 y - 1)$

$= (x^{2} + 2) (4 x^{2} - 1)$ ... $[∵ y = x^{2}]$

$= (x^{2} + 2) [{(2 x)}^{2} - 1^{2}]$

$= (x^{2} + 2) [(2 x - 1) (2 x + 1)]$ ... $[∵ a^{2} - b^{2} = (a - b) (a + b)]$

$= (x^{2} + 2) (2 x - 1) (2 x + 1)$

The question number 66 of Exercise 3C RS Aggarwal Class 9 asks us to factorise the given algebraic expression.

66. Evaluate $[{(999)}^{2} - 1]$

Answer

We have
$[{(999)}^{2} - 1]$

$= (999 - 1) (999 + 1)$ ... $[∵ a^{2} - b^{2} = (a - b) (a + b)]$

$= 998 \times 1000$

$= 998000$

Exercise 3C RS Aggarwal Class 9 Mathematics Solutions

Factorise:

1. x2+11x+30

Answer

2. x2+18x+32

Answer

3. x2+20x−69

Answer

4. x2+19x−150

Answer

5. x2+7x−98

Answer

6. x2+23x−24

Answer

7. x2−21x+90

Answer

8. x2−22x+120

Answer

9. x2−4x+3

Answer

10. x2+76x+60

Answer

11. x2+33x+6

Answer

12. x2+66x+48

Answer

13. x2+55x+30

Answer

14. x2−24x−180

Answer

15. x2−32x−105

Answer

16. x2−11x−80

Answer

17. 6−x−x2

Answer

18. x2−3x−6

Answer

19. 40+3x−x2

Answer

20. x2−26x+133

Answer

21. x2−23x−24

Answer

22. x2−35x−20

Answer

23. x2+2x−24

Answer

24. x2−22x−30

Answer

25. x2−x−156

Answer

26. x2−32x−105

Answer

27. 9x2+18x+8

Answer

28. 6x2+17x+12

Answer

29. 18x2+3x−10

Answer

30. 2x2+11x−21

Answer

31. 15x2+2x−8

Answer

32. 21x2+5x−6

Answer

33. 24x2−41x+12

Answer

34. 3x2−14x+8

Answer

35. 2x2+3x−90

Answer

36. 5x2+2x−35

Answer

37. 23x2+x−53

Answer

38. 7x2+214x+2

Answer

39. 63x2−47x+53

Answer

1. $x^{2} + 11 x + 30$

2. $x^{2} + 18 x + 32$

3. $x^{2} + 20 x - 69$

4. $x^{2} + 19 x - 150$

5. $x^{2} + 7 x - 98$

6. $x^{2} + 2 \sqrt{3} x - 24$

7. $x^{2} - 21 x + 90$

8. $x^{2} - 22 x + 120$

9. $x^{2} - 4 x + 3$

10. $x^{2} + 7 \sqrt{6} x + 60$

11. $x^{2} + 3 \sqrt{3} x + 6$

12. $x^{2} + 6 \sqrt{6} x + 48$

13. $x^{2} + 5 \sqrt{5} x + 30$

14. $x^{2} - 24 x - 180$

15. $x^{2} - 32 x - 105$

16. $x^{2} - 11 x - 80$

17. $6 - x - x^{2}$

18. $x^{2} - \sqrt{3} x - 6$

19. $40 + 3 x - x^{2}$

20. $x^{2} - 26 x + 133$

21. $x^{2} - 2 \sqrt{3} x - 24$

22. $x^{2} - 3 \sqrt{5} x - 20$

23. $x^{2} + \sqrt{2} x - 24$

24. $x^{2} - 2 \sqrt{2} x - 30$

25. $x^{2} - x - 156$

26. $x^{2} - 32 x - 105$

27. $9 x^{2} + 18 x + 8$

28. $6 x^{2} + 17 x + 12$

29. $18 x^{2} + 3 x - 10$

30. $2 x^{2} + 11 x - 21$

31. $15 x^{2} + 2 x - 8$

32. $21 x^{2} + 5 x - 6$

33. $24 x^{2} - 41 x + 12$

34. $3 x^{2} - 14 x + 8$

35. $2 x^{2} + 3 x - 90$

36. $\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$

37. $2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$

38. $7 x^{2} + 2 \sqrt{14} x + 2$

39. $6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

40. $5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

41. $\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

42. $\sqrt{2} x^{2} + 3 x + \sqrt{2}$

43. $2 x^{2} + 3 \sqrt{3} x + 3$

44. $15 x^{2} - x - 28$

45. $6 x^{2} - 5 x - 21$

46. $2 x^{2} - 7 x - 15$

47. $5 x^{2} - 16 x - 21$

48. $6 x^{2} - 11 x - 35$

49. $9 x^{2} - 3 x - 20$

50. $10 x^{2} - 9 x - 7$

51. $x^{2} - 2 x + \frac{7}{16}$

52. $\frac{1}{3} x^{2} - 2 x - 9$

53. $x^{2} + \frac{12}{35} x + \frac{1}{35}$

54. $21 x^{2} - 2 x + \frac{1}{21}$

55. $\frac{3}{2} x^{2} + 16 x + 10$

56. $\frac{2}{3} x^{2} - \frac{17}{3} x - 28$

57. $\frac{3}{5} x^{2} - \frac{19}{5} x + 4$

58. $2 x^{2} - x + \frac{1}{8}$

59. $2 {(x + y)}^{2} - 9 (x + y) - 5$

60. $9 {(2 a - b)}^{2} - 4 (2 a - b) - 13$

61. $7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$

62. $10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$

63. $6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$

64. ${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$

65. $4 x^{4} + 7 x^{2} - 2$

66. Evaluate $[{(999)}^{2} - 1]$