myhelper: RS AGGARWAL CLASS 9 Chapter 3 FACTORISATION OF POLYNOMIAL EXERCISE 3C

EXERCISE 3C

PAGE NO-114

Question 1:

Factorize:
x² + 11x + 30

Answer 1:

We have:
$x^{2} + 11 x + 30$
We have to split 11 into two numbers such that their sum of is 11 and their product is 30.
Clearly, $5 + 6 = 11 and 5 \times 6 = 30$ .

$∴ x^{2} + 11 x + 30 = x^{2} + 5 x + 6 x + 30 = x (x + 5) + 6 (x + 5) = (x + 5) (x + 6)$

Question 2:

Factorize:
x² + 18x + 32

Answer 2:

We have:
$x^{2} + 18 x + 32$
We have to split 18 into two numbers such that their sum is 18 and their product is 32.
Clearly, $16 + 2 = 18 and 16 \times 2 = 32$ .

$∴ x^{2} + 18 x + 32 = x^{2} + 16 x + 2 x + 32 = x (x + 16) + 2 (x + 16) = (x + 16) (x + 2)$

Question 3:

Factorise:
x² + 20x – 69

Answer 3:

$x^{2} + 20 x - 69 = x^{2} + 23 x - 3 x - 69 = x (x + 23) - 3 (x + 23) = (x + 23) (x - 3)$

Question 4:

x² + 19x – 150

Answer 4:

$x^{2} + 19 x - 150 = x^{2} + 25 x - 6 x - 150 = x (x + 25) - 6 (x + 25) = (x + 25) (x - 6)$

Question 5:

Factorise:
x² + 7x – 98

Answer 5:

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

Question 6:

Factorise:
$x^{2} + 2 \sqrt{3} x - 24$

Answer 6:

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

Question 7:

Factorise:
x² – 21x + 90

Answer 7:

$x^{2} - 21 x + 90 = x^{2} - 15 x - 6 x + 90 = x (x - 15) - 6 (x - 15) = (x - 6) (x - 15)$

Question 8:

Factorise:
x² – 22x + 120

Answer 8:

$x^{2} - 22 x + 120 = x^{2} - 12 x - 10 x + 120 = x (x - 12) - 10 (x - 12) = (x - 10) (x - 12)$

Question 9:

Factorise:
x² – 4x + 3

Answer 9:

$x^{2} - 4 x + 3 = x^{2} - 3 x - x + 3 = x (x - 3) - 1 (x - 3) = (x - 1) (x - 3)$

Question 10:

Factorise:
$x^{2} + 7 \sqrt{6} x + 60$

Answer 10:

$x^{2} + 7 \sqrt{6} x + 60 = x^{2} + 5 \sqrt{6} x + 2 \sqrt{6} x + 60 = x (x + 5 \sqrt{6}) + 2 \sqrt{6} (x + 5 \sqrt{6}) = (x + 5 \sqrt{6}) (x + 2 \sqrt{6})$

Question 11:

Factorise:
$x^{2} + 3 \sqrt{3} x + 6$

Answer 11:

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

Question 12:

Factorise:
$x^{2} + 6 \sqrt{6} x + 48$

Answer 12:

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

Question 13:

Factorise:
$x^{2} + 5 \sqrt{5} x + 30$

Answer 13:

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

Question 14:

Factorise:
$x^{2} - 24 x - 180$

Answer 14:

$x^{2} - 24 x - 180 = x^{2} - 30 x + 6 x - 180 = x (x - 30) + 6 (x - 30) = (x - 30) (x + 6)$

Question 15:

Factorise:
x² – 32x – 105

Answer 15:

$x^{2} - 32 x - 105 = x^{2} - 35 x + 3 x - 105 = x (x - 35) + 3 (x - 35) = (x - 35) (x + 3)$

Question 16:

Factorise:
x² – 11x – 80

Answer 16:

$x^{2} - 11 x - 80 = x^{2} - 16 x + 5 x - 80 = x (x - 16) + 5 (x - 16) = (x - 16) (x + 5)$

Question 17:

Factorise:
6 – x – x²

Answer 17:

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

Question 18:

Factorise:
$x^{2} - \sqrt{3} x - 6$

Answer 18:

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

Question 19:

Factorise:
40 + 3x – x²

Answer 19:

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

Question 20:

Factorise:
x² – 26x + 133

Answer 20:

$x^{2} - 26 x + 133 = x^{2} - 19 x - 7 x + 133 = x (x - 19) - 7 (x - 19) = (x - 19) (x - 7)$

Question 21:

Factorise:
$x^{2} - 2 \sqrt{3} x - 24$

Answer 21:

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

Question 22:

Factorise:
$x^{2} - 3 \sqrt{5} x - 20$

Answer 22:

$x^{2} - 3 \sqrt{5} x - 20 = x^{2} - 4 \sqrt{5} x + \sqrt{5} x - 20 = x (x - 4 \sqrt{5}) + \sqrt{5} (x - 4 \sqrt{5}) = (x - 4 \sqrt{5}) (x + \sqrt{5})$

Question 23:

Factorise:
$x^{2} + \sqrt{2} x - 24$

Answer 23:

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

Question 24:

Factorise:
$x^{2} - 2 \sqrt{2} x - 30$

Answer 24:

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

Question 25:

Factorize:
x² − x − 156

Answer 25:

We have:
$x^{2} - x - 156$
We have to split ( $-$ 1) into two numbers such that their sum is ( $-$ 1) and their product is ( $-$ 156).
Clearly, $- 13 + 12 = - 1 and - 13 \times 12 = - 156$ .

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

Question 26:

Factorise:
x² – 32x – 105

Answer 26:

$x^{2} - 32 x - 105 = x^{2} - 35 x + 3 x - 105 = x (x - 35) + 3 (x - 35) = (x - 35) (x + 3)$

Question 27:

Factorise:
9x²+ 18x + 8

Answer 27:

$9 x^{2} + 18 x + 8 = 9 x^{2} + 12 x + 6 x + 8 = 3 x (3 x + 4) + 2 (3 x + 4) = (3 x + 4) (3 x + 2)$

Question 28:

Factorise:
6x²+ 17x + 12

Answer 28:

$6 x^{2} + 17 x + 12 = 6 x^{2} + 9 x + 8 x + 12 = 3 x (2 x + 3) + 4 (2 x + 3) = (2 x + 3) (3 x + 4)$

Question 29:

Factorize:
18x² + 3x − 10

Answer 29:

We have:
$18 x^{2} + 3 x - 10$
We have to split 3 into two numbers such that their sum is 3 and their product is ( $-$ 180), i.e., $18 \times (- 10)$ .
Clearly, $15 + (- 12) = 3 and 15 \times (- 12) = - 180$ .

$∴ 18 x^{2} + 3 x - 10 = 18 x^{2} + 15 x - 12 x - 10 = 3 x (6 x + 5) - 2 (6 x + 5) = (6 x + 5) (3 x - 2)$

Question 30:

Factorize:
2x² + 11x − 21

Answer 30:

We have:
$2 x^{2} + 11 x - 21$
We have to split 11 into two numbers such that their sum is 11 and their product is ( $-$ 42), i.e., $2 \times (- 21)$ .
Clearly, $14 + (- 3) = 11 and 14 \times (- 3) = - 42$ .

$∴ 2 x^{2} + 11 x - 21 = 2 x^{2} + 14 x - 3 x - 21 = 2 x (x + 7) - 3 (x + 7) = (x + 7) (2 x - 3)$

Question 31:

Factorize:
15x² + 2x − 8

Answer 31:

We have:
$15 x^{2} + 2 x - 8$
We have to split 2 into two numbers such that their sum is 2 and their product is ( $-$ 120), i.e., $15 \times (- 8)$ .
Clearly, $12 + (- 10) = 2 and 12 \times (- 10) = - 120$ .

$∴ 15 x^{2} + 2 x - 8 = 15 x^{2} + 12 x - 10 x - 8 = 3 x (5 x + 4) - 2 (5 x + 4) = (5 x + 4) (3 x - 2)$

Question 32:

Factorise:
21x²+ 5x – 6

Answer 32:

$21 x^{2} + 5 x - 6 = 21 x^{2} + 14 x - 9 x - 6 = 7 x (3 x + 2) - 3 (3 x + 2) = (3 x + 2) (7 x - 3)$

Question 33:

Factorize:
24x² − 41x + 12

Answer 33:

We have:
$24 x^{2} - 41 x + 12$
We have to split ( $-$ 41) into two numbers such that their sum is ( $-$ 41) and their product is 288, i.e., $24 \times 12$ .
Clearly, $(- 32) + (- 9) = - 41 and (- 32) \times (- 9) = 288$ .

$∴ 24 x^{2} - 41 x + 12 = 24 x^{2} - 32 x - 9 x + 12 = 8 x (3 x - 4) - 3 (3 x - 4) = (3 x - 4) (8 x - 3)$

Question 34:

Factorise:
3x²– 14x + 8

Answer 34:

$3 x^{2} - 14 x + 8 = 3 x^{2} - 12 x - 2 x + 8 = 3 x (x - 4) - 2 (x - 4) = (x - 4) (3 x - 2)$

Hence, factorisation of 3x²– 14x + 8 is $(x - 4) (3 x - 2)$ .

Question 35:

Factorize:
2x² + 3x − 90

Answer 35:

We have:
$2 x^{2} + 3 x - 90$
We have to split 3 into two numbers such that their sum is 3 and their product is ( $-$ 180), i.e., $2 \times (- 90)$ .
Clearly, $- 12 + 15 = 3 and - 12 \times 15 = - 180$ .

$∴ 2 x^{2} + 3 x - 90 = 2 x^{2} - 12 x + 15 x - 90 = 2 x (x - 6) + 15 (x - 6) = (x - 6) (2 x + 15)$

Question 36:

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

Answer 36:

We have:
$\sqrt{5} x^{2} + 2 x - 3 \sqrt{5}$
We have to split 2 into two numbers such that their sum is 2 and product is ( $-$ 15), i.e., $\sqrt{5} \times (- 3 \sqrt{5})$ .
Clearly, $5 + (- 3) = 2 and 5 \times (- 3) = - 15$ .

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

Question 37:

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

Answer 37:

We have:
$2 \sqrt{3} x^{2} + x - 5 \sqrt{3}$
We have to split 1 into two numbers such that their sum is 1 and product is 30, i.e., $2 \sqrt{3} \times (- 5 \sqrt{3})$ .
Clearly, $6 + (- 5) = 1 and 6 \times (- 5) = - 30$ .

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

Question 38:

Factorize:
$7 x^{2} + 2 \sqrt{14} x + 2$

Answer 38:

We have:
$7 x^{2} + 2 \sqrt{14} x + 2$
We have to split $2 \sqrt{14}$ into two numbers such that their sum is $2 \sqrt{14}$ and product is 14.
Clearly, $\sqrt{14} + \sqrt{14} = 2 \sqrt{14} and \sqrt{14} \times \sqrt{14} = 14$ .
$∴ 7 x^{2} + 2 \sqrt{14} x + 2 = 7 x^{2} + \sqrt{14} x + \sqrt{14} x + 2 = \sqrt{7} x (\sqrt{7} x + \sqrt{2}) + \sqrt{2} (\sqrt{7} x + \sqrt{2}) = (\sqrt{7} x + \sqrt{2}) (\sqrt{7} x + \sqrt{2}) = {(\sqrt{7} x + \sqrt{2})}^{2}$

Question 39:

Factorize:
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$

Answer 39:

We have:
$6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3}$
Now, we have to split ( $-$ 47) into two numbers such that their sum is ( $-$ 47) and their product is 90.
Clearly, $(- 45) + (- 2) = - 47 and (- 45) \times (- 2) = 90$ .

$∴ 6 \sqrt{3} x^{2} - 47 x + 5 \sqrt{3} = 6 \sqrt{3} x^{2} - 2 x - 45 x + 5 \sqrt{3} = 2 x (3 \sqrt{3} x - 1) - 5 \sqrt{3} (3 \sqrt{3} x - 1) = (3 \sqrt{3} x - 1) (2 x - 5 \sqrt{3})$

Question 40:

Factorize:
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$

Answer 40:

We have:
$5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5}$
We have to split 20 into two numbers such that their sum is 20 and their product is 75.
Clearly,
$15 + 5 = 20 and 15 \times 5 = 75$

$∴ 5 \sqrt{5} x^{2} + 20 x + 3 \sqrt{5} = 5 \sqrt{5} x^{2} + 15 x + 5 x + 3 \sqrt{5} = 5 x (\sqrt{5} x + 3) + \sqrt{5} (\sqrt{5} x + 3) = (\sqrt{5} x + 3) (5 x + \sqrt{5})$

Question 41:

Factorise:
$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$

Answer 41:

$\sqrt{3} x^{2} + 10 x + 8 \sqrt{3} = \sqrt{3} x^{2} + 6 x + 4 x + 8 \sqrt{3} = \sqrt{3} x (x + 2 \sqrt{3}) + 4 (x + 2 \sqrt{3}) = (x + 2 \sqrt{3}) (\sqrt{3} x + 4)$

Hence, factorisation of $\sqrt{3} x^{2} + 10 x + 8 \sqrt{3}$ is $(x + 2 \sqrt{3}) (\sqrt{3} x + 4)$ .

Question 42:

Factorize:
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$

Answer 42:

We have:
$\sqrt{2} x^{2} + 3 x + \sqrt{2}$
We have to split 3 into two numbers such that their sum is 3 and their product is 2, i.e., $\sqrt{2} \times \sqrt{2}$ .
Clearly, $2 + 1 = 3 and 2 \times 1 = 2$ .

$∴ \sqrt{2} x^{2} + 3 x + \sqrt{2} = \sqrt{2} x^{2} + 2 x + x + \sqrt{2} = \sqrt{2} x (x + \sqrt{2}) + 1 (x + \sqrt{2}) = (x + \sqrt{2}) (\sqrt{2} x + 1)$

Question 43:

Factorize:
$2 x^{2} + 3 \sqrt{3} x + 3$

Answer 43:

We have:
$2 x^{2} + 3 \sqrt{3} x + 3$
We have to split $3 \sqrt{3}$ into two numbers such that their sum is $3 \sqrt{3}$ and their product is 6, i.e., $2 \times 3$ .
Clearly, $2 \sqrt{3} + \sqrt{3} = 3 \sqrt{3} and 2 \sqrt{3} \times \sqrt{3} = 6$ .

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

Question 44:

Factorize:
15x² − x − 128

Answer 44:

We have:
$15 x^{2} - x - 28$
We have to split ( $-$ 1) into two numbers such that their sum is ( $-$ 1) and their product is ( $-$ 420), i.e., $15 \times (- 28)$ .
Clearly, $(- 21) + 20 = - 1 and (- 21) \times 20 = - 420$ .

$∴ 15 x^{2} - x - 28 = 15 x^{2} - 21 x + 20 x - 28 = 3 x (5 x - 7) + 4 (5 x - 7) = (5 x - 7) (3 x + 4)$

Question 45:

Factorize:
6x² − 5x − 21

Answer 45:

We have:
$6 x^{2} - 5 x - 21$
We have to split ( $-$ 5) into two numbers such that their sum is ( $-$ 5) and their product is ( $-$ 126), i.e., $6 \times (- 21)$ .
Clearly, $9 + (- 14) = - 5 and 9 \times (- 14) = - 126$ .

$∴ 6 x^{2} - 5 x - 21 = 6 x^{2} + 9 x - 14 x - 21 = 3 x (2 x + 3) - 7 (2 x + 3) = (2 x + 3) (3 x - 7)$

Question 46:

Factorize:
2x² − 7x − 15

Answer 46:

We have:
$2 x^{2} - 7 x - 15$
We have to split ( $-$ 7) into two numbers such that their sum is ( $-$ 7) and their product is ( $-$ 30), i.e., $2 \times (- 15)$ .
Clearly, $(- 10) + 3 = - 7 and (- 10) \times 3 = - 30$ .

$∴ 2 x^{2} - 7 x - 15 = 2 x^{2} - 10 x + 3 x - 15 = 2 x (x - 5) + 3 (x - 5) = (x - 5) (2 x + 3)$

Question 47:

Factorize:
5x² − 16x − 21

Answer 47:

We have:
$5 x^{2} - 16 x - 21$
We have to split ( $-$ 16) into two numbers such that their sum is ( $-$ 16) and their product is ( $-$ 105), i.e., $5 \times (- 21)$ .
Clearly, $(- 21) + 5 = - 16 and (- 21) \times 5 = - 105$ .

$∴ 5 x^{2} - 16 x - 21 = 5 x^{2} + 5 x - 21 x - 21 = 5 x (x + 1) - 21 (x + 1) = (x + 1) (5 x - 21)$

Question 48:

Factorise:
6x² – 11x – 35

Answer 48:

$6 x^{2} - 11 x - 35 = 6 x^{2} - 21 x + 10 x - 35 = 3 x (2 x - 7) + 5 (2 x - 7) = (2 x - 7) (3 x + 5)$

Hence, factorisation of 6x² – 11x – 35 is $(2 x - 7) (3 x + 5)$ .

Question 49:

Factorise:
9x² – 3x – 20

Answer 49:

$9 x^{2} - 3 x - 20 = 9 x^{2} - 15 x + 12 x - 20 = 3 x (3 x - 5) + 4 (3 x - 5) = (3 x - 5) (3 x + 4)$

Hence, factorisation of 9x² – 3x – 20 is $(3 x - 5) (3 x + 4)$ .

Question 50:

Factorize:
10x² − 9x − 7

Answer 50:

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

We have to split ( $-$ 9) into two numbers such that their sum is ( $-$ 9) and their product is ( $-$ 70), i.e., $10 \times (- 7)$ .
Clearly, $(- 14) + 5 = - 9 and (- 14) \times 5 = - 70$ .

$∴ 10 x^{2} - 9 x - 7 = 10 x^{2} + 5 x - 14 x - 7 = 5 x (2 x + 1) - 7 (2 x + 1) = (2 x + 1) (5 x - 7)$

Question 51:

Factorize:
$x^{2} - 2 x + \frac{7}{16}$

Answer 51:

$We have : x^{2} - 2 x + \frac{7}{16} = \frac{16 x^{2} - 32 x + 7}{16} = \frac{1}{16} (16 x^{2} - 32 x + 7)$

Now, we have to split ( $-$ 32) into two numbers such that their sum is ( $-$ 32) and their product is 112, i.e., $16 \times 7$ .
Clearly, $(- 4) + (- 28) = - 32 and (- 4) \times (- 28) = 112$ .

$∴ x^{2} - 2 x + \frac{7}{16} = \frac{1}{16} (16 x^{2} - 32 x + 7) = \frac{1}{16} (16 x^{2} - 4 x - 28 x + 7) = \frac{1}{16} [4 x (4 x - 1) - 7 (4 x - 1)] = \frac{1}{16} (4 x - 1) (4 x - 7)$

Question 52:

Factorise:
$\frac{1}{3} x^{2} - 2 x - 9$

Answer 52:

$\frac{1}{3} x^{2} - 2 x - 9 = \frac{x^{2} - 6 x - 27}{3} = \frac{x^{2} - 9 x + 3 x - 27}{3} = \frac{x (x - 9) + 3 (x - 9)}{3} = \frac{(x - 9) (x + 3)}{3} = \frac{(x - 9)}{3} \times \frac{(x + 3)}{1} = (\frac{1}{3} x - 3) (x + 3)$

Hence, factorisation of $\frac{1}{3} x^{2} - 2 x - 9$ is $(\frac{1}{3} x - 3) (x + 3)$ .

Question 53:

Factorise:
$x^{2} + \frac{12}{35} x + \frac{1}{35}$

Answer 53:

$x^{2} + \frac{12}{35} x + \frac{1}{35} = \frac{35 x^{2} + 12 x + 1}{35} = \frac{35 x^{2} + 7 x + 5 x + 1}{35} = \frac{7 x (5 x + 1) + 1 (5 x + 1)}{35} = \frac{(5 x + 1) (7 x + 1)}{35} = \frac{(5 x + 1) (7 x + 1)}{5 \times 7} = \frac{(5 x + 1)}{5} \times \frac{(7 x + 1)}{7} = (x + \frac{1}{5}) (x + \frac{1}{7})$

Hence, factorisation of $x^{2} + \frac{12}{35} x + \frac{1}{35}$ is $(x + \frac{1}{5}) (x + \frac{1}{7})$ .

Question 54:

Factorise:
$21 x^{2} - 2 x + \frac{1}{21}$

Answer 54:

$21 x^{2} - 2 x + \frac{1}{21} = 21 x^{2} - x - x + \frac{1}{21} = 21 x (x - \frac{1}{21}) - 1 (x - \frac{1}{21}) = (x - \frac{1}{21}) (21 x - 1)$

Hence, factorisation of $21 x^{2} - 2 x + \frac{1}{21}$ is $(x - \frac{1}{21}) (21 x - 1)$ .

Question 55:

Factorise:
$\frac{3}{2} x^{2} + 16 x + 10$

Answer 55:

$\frac{3}{2} x^{2} + 16 x + 10 = \frac{3}{2} x^{2} + 15 x + x + 10 = 3 x (\frac{1}{2} x + 5) + 1 (x + 10) = \frac{3}{2} x (x + 10) + 1 (x + 10) = (x + 10) (\frac{3}{2} x + 1)$

Hence, factorisation of $\frac{3}{2} x^{2} + 16 x + 10$ is $(x + 10) (\frac{3}{2} x + 1)$ .

Question 56:

Factorise:
$\frac{2}{3} x^{2} - \frac{17}{3} x - 28$

Answer 56:

$\frac{2}{3} x^{2} - \frac{17}{3} x - 28 = \frac{2}{3} x^{2} - 8 x + \frac{7}{3} x - 28 = 2 x (\frac{1}{3} x - 4) + 7 (\frac{1}{3} x - 4) = (\frac{1}{3} x - 4) (2 x + 7)$

Hence, factorisation of $\frac{2}{3} x^{2} - \frac{17}{3} x - 28$ is $(\frac{1}{3} x - 4) (2 x + 7)$ .

Question 57:

Factorise:
$\frac{3}{5} x^{2} - \frac{19}{5} x + 4$

Answer 57:

$\frac{3}{5} x^{2} - \frac{19}{5} x + 4 = \frac{3}{5} x^{2} - 3 x - \frac{4}{5} x + 4 = 3 x (\frac{1}{5} x - 1) - 4 (\frac{1}{5} x - 1) = (\frac{1}{5} x - 1) (3 x - 4)$

Hence, factorisation of $\frac{3}{5} x^{2} - \frac{19}{5} x + 4$ is $(\frac{1}{5} x - 1) (3 x - 4)$ .

Question 58:

Factorise:
$2 x^{2} - x + \frac{1}{8}$

Answer 58:

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

Hence, factorisation of $2 x^{2} - x + \frac{1}{8}$ is $(x - \frac{1}{4}) (2 x - \frac{1}{2})$ .

Question 59:

Factorize:
2(x + y)² − 9(x + y) − 5

Answer 59:

We have:
$2 {(x + y)}^{2} - 9 (x + y) - 5 Let : (x + y) = u$
Thus, the given expression becomes
$2 u^{2} - 9 u - 5$
Now, we have to split ( $-$ 9) into two numbers such that their sum is ( $-$ 9) and their product is ( $-$ 10).
Clearly, $- 10 + 1 = - 9 and - 10 \times 1 = - 10$ .

$∴ 2 u^{2} - 9 u - 5 = 2 u^{2} - 10 u + u - 5 = 2 u (u - 5) + 1 (u - 5) = (u - 5) (2 u + 1)$
Putting $u = (x + y)$ , we get:
$2 {(x + y)}^{2} - 9 (x + y) - 5 = (x + y - 5) [2 (x + y) + 1] = (x + y - 5) (2 x + 2 y + 1)$

Question 60:

Factorize:
9(2a − b)² − 4(2a − b) − 13

Answer 60:

We have:
$9 (2 a - b)^{2} - 4 (2 a - b) - 13 Let : (2 a - b) = p$
Thus, the given expression becomes
$9 p^{2} - 4 p - 13$
Now, we must split ( $-$ 4) into two numbers such that their sum is ( $-$ 4) and their product is ( $-$ 117).
Clearly, $- 13 + 9 = - 4 and - 13 \times 9 = - 117$ .
$∴ 9 p^{2} - 4 p - 13 = 9 p^{2} + 9 p - 13 p - 13 = 9 p (p + 1) - 13 (p + 1) = (p + 1) (9 p - 13)$
Putting $p = (2 a - b)$ , we get:
$9 {(2 a - b)}^{2} - 4 (2 a - b) - 13 = [(2 a - b) + 1] [9 (2 a - b) - 13] = (2 a - b + 1) [18 a - 9 b - 13]$

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Question 61:

Factorise:
$7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$

Answer 61:

$7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12 = 7 {(x - 2 y)}^{2} - 21 (x - 2 y) - 4 (x - 2 y) + 12 = [7 (x - 2 y)] (x - 2 y - 3) - 4 (x - 2 y - 3) = [7 (x - 2 y) - 4] (x - 2 y - 3) = (7 x - 14 y - 4) (x - 2 y - 3)$

Hence, factorisation of $7 {(x - 2 y)}^{2} - 25 (x - 2 y) + 12$ is $(7 x - 14 y - 4) (x - 2 y - 3)$ .

Question 62:

Factorise:
$10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$

Answer 62:

$10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3 = 10 {(3 x + \frac{1}{x})}^{2} - 6 (3 x + \frac{1}{x}) + 5 (3 x + \frac{1}{x}) - 3 = [2 (3 x + \frac{1}{x})] [5 (3 x + \frac{1}{x}) - 3] + 1 [5 (3 x + \frac{1}{x}) - 3] = [5 (3 x + \frac{1}{x}) - 3] [2 (3 x + \frac{1}{x}) + 1] = (15 x + \frac{5}{x} - 3) (6 x + \frac{2}{x} + 1)$

Hence, factorisation of $10 {(3 x + \frac{1}{x})}^{2} - (3 x + \frac{1}{x}) - 3$ is $(15 x + \frac{5}{x} - 3) (6 x + \frac{2}{x} + 1)$ .

Question 63:

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

Answer 63:

$6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20 = 6 {(2 x - \frac{3}{x})}^{2} + 15 (2 x - \frac{3}{x}) - 8 (2 x - \frac{3}{x}) - 20 = [3 (2 x - \frac{3}{x})] [2 (2 x - \frac{3}{x}) + 5] - 4 [2 (2 x - \frac{3}{x}) + 5] = [2 (2 x - \frac{3}{x}) + 5] [3 (2 x - \frac{3}{x}) - 4] = (4 x - \frac{6}{x} + 5) (6 x - \frac{9}{x} - 4)$

Hence, factorisation of $6 {(2 x - \frac{3}{x})}^{2} + 7 (2 x - \frac{3}{x}) - 20$ is $(4 x - \frac{6}{x} + 5) (6 x - \frac{9}{x} - 4)$ .

Question 64:

Factorise:
${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$

Answer 64:

${(a + 2 b)}^{2} + 101 (a + 2 b) + 100 = {(a + 2 b)}^{2} + 100 (a + 2 b) + 1 (a + 2 b) + 100 = (a + 2 b) [(a + 2 b) + 100] + 1 [(a + 2 b) + 100] = [(a + 2 b) + 1] [(a + 2 b) + 100] = (a + 2 b + 1) (a + 2 b + 100)$

Hence, factorisation of ${(a + 2 b)}^{2} + 101 (a + 2 b) + 100$ is $(a + 2 b + 1) (a + 2 b + 100)$ .

Question 65:

Factorise:
4x⁴ + 7x² – 2

Answer 65:

$4 x^{4} + 7 x^{2} - 2 = 4 x^{4} + 8 x^{2} - x^{2} - 2 = 4 x^{2} (x^{2} + 2) - 1 (x^{2} + 2) = (4 x^{2} - 1) (x^{2} + 2)$

Hence, factorisation of 4x⁴ + 7x² – 2 is $(4 x^{2} - 1) (x^{2} + 2)$ .

Question 66:

Evaluate {(999)2 – 1}.

Answer 66:

$\{{(999)}^{2} - 1\} = \{{(999)}^{2} - 1^{2}\} = (999 - 1) (999 + 1) = (998) (1000) = 998000$

Hence, {(999)² – 1} = 998000.