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

$$\begin{gathered} \therefore x^2+11x+30 = x^2+5x+6x+30 \\ = x(x+5)+6(x+5) \\ =(x+5)(x+6) \end{gathered}$$

Question 2:

Factorize:
x² + 18x + 32

Answer 2:

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

$$\begin{gathered} \therefore x^2+18x+32=x^2+16x+2x+32 \\ =x(x+16)+2(x+16) \\ =(x+16)(x+2) \end{gathered}$$

Question 3:

Factorise:
x² + 20x – 69

Answer 3:

$$\begin{gathered} x^2+20x-69 \\ =x^2+23x-3x-69 \\ =x\left(x+23\right)-3\left(x+23\right) \\ =\left(x+23\right)\left(x-3\right) \end{gathered}$$

Question 4:

x² + 19x – 150

Answer 4:

$$\begin{gathered} x^2+19x-150 \\ =x^2+25x-6x-150 \\ =x\left(x+25\right)-6\left(x+25\right) \\ =\left(x+25\right)\left(x-6\right) \end{gathered}$$

Question 5:

Factorise:
x² + 7x – 98

Answer 5:

$$\begin{gathered} x^2+7x-98 \\ =x^2+14x-7x-98 \\ =x\left(x+14\right)-7\left(x+14\right) \\ =\left(x+14\right)\left(x-7\right) \end{gathered}$$

Question 6:

Factorise:
$x^2+2\sqrt{3}x–24$

Answer 6:

$$\begin{gathered} x^2+2\sqrt{3}x–24 \\ =x^2+4\sqrt{3}x-2\sqrt{3}x-24 \\ =x\left(x+4\sqrt{3}\right)-2\sqrt{3}\left(x+4\sqrt{3}\right) \\ =\left(x+4\sqrt{3}\right)\left(x-2\sqrt{3}\right) \end{gathered}$$

Question 7:

Factorise:
x² – 21x + 90

Answer 7:

$$\begin{gathered} x^2-21x+90 \\ =x^2-15x-6x+90 \\ =x\left(x-15\right)-6\left(x-15\right) \\ =\left(x-6\right)\left(x-15\right) \end{gathered}$$

Question 8:

Factorise:
x² – 22x + 120

Answer 8:

$$\begin{gathered} x^2-22x+120 \\ =x^2-12x-10x+120 \\ =x\left(x-12\right)-10\left(x-12\right) \\ =\left(x-10\right)\left(x-12\right) \end{gathered}$$

Question 9:

Factorise:
x² – 4x + 3

Answer 9:

$$\begin{gathered} x^2-4x+3 \\ =x^2-3x-x+3 \\ =x\left(x-3\right)-1\left(x-3\right) \\ =\left(x-1\right)\left(x-3\right) \end{gathered}$$

Question 10:

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

Answer 10:

$$\begin{gathered} x^2+7\sqrt{6}x+60 \\ =x^2+5\sqrt{6}x+2\sqrt{6}x+60 \\ =x\left(x+5\sqrt{6}\right)+2\sqrt{6}\left(x+5\sqrt{6}\right) \\ =\left(x+5\sqrt{6}\right)\left(x+2\sqrt{6}\right) \end{gathered}$$

Question 11:

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

Answer 11:

$$\begin{gathered} x^2+3\sqrt{3}x+6 \\ =x^2+2\sqrt{3}x+\sqrt{3}x+6 \\ =x\left(x+2\sqrt{3}\right)+\sqrt{3}\left(x+2\sqrt{3}\right) \\ =\left(x+2\sqrt{3}\right)\left(x+\sqrt{3}\right) \end{gathered}$$

Question 12:

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

Answer 12:

$$\begin{gathered} x^2+6\sqrt{6}x+48 \\ =x^2+4\sqrt{6}x+2\sqrt{6}x+48 \\ =x\left(x+4\sqrt{6}\right)+2\sqrt{6}\left(x+4\sqrt{6}\right) \\ =\left(x+4\sqrt{6}\right)\left(x+2\sqrt{6}\right) \end{gathered}$$

Question 13:

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

Answer 13:

$$\begin{gathered} x^2+5\sqrt{5}x+30 \\ =x^2+3\sqrt{5}x+2\sqrt{5}x+30 \\ =x\left(x+3\sqrt{5}\right)+2\sqrt{5}\left(x+3\sqrt{5}\right) \\ =\left(x+3\sqrt{5}\right)\left(x+2\sqrt{5}\right) \end{gathered}$$

Question 14:

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

Answer 14:

$$\begin{gathered} x^2-24x-180 \\ =x^2-30x+6x-180 \\ =x\left(x-30\right)+6\left(x-30\right) \\ =\left(x-30\right)\left(x+6\right) \end{gathered}$$

Question 15:

Factorise:
x² – 32x – 105

Answer 15:

$$\begin{gathered} x^2-32x-105 \\ =x^2-35x+3x-105 \\ =x\left(x-35\right)+3\left(x-35\right) \\ =\left(x-35\right)\left(x+3\right) \end{gathered}$$

Question 16:

Factorise:
x² – 11x – 80

Answer 16:

$$\begin{gathered} x^2-11x-80 \\ =x^2-16x+5x-80 \\ =x\left(x-16\right)+5\left(x-16\right) \\ =\left(x-16\right)\left(x+5\right) \end{gathered}$$

Question 17:

Factorise:
6 – x – x²

Answer 17:

$$\begin{gathered} -x^2-x+6 \\ =-x^2-3x+2x+6 \\ =-x\left(x+3\right)+2\left(x+3\right) \\ =\left(x+3\right)\left(-x+2\right) \\ =\left(x+3\right)\left(2-x\right) \end{gathered}$$

Question 18:

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

Answer 18:

$$\begin{gathered} x^2-\sqrt{3}x-6 \\ =x^2-2\sqrt{3}x+\sqrt{3}x-6 \\ =x\left(x-2\sqrt{3}\right)+\sqrt{3}\left(x-2\sqrt{3}\right) \\ =\left(x-2\sqrt{3}\right)\left(x+\sqrt{3}\right) \end{gathered}$$

Question 19:

Factorise:
40 + 3x – x²

Answer 19:

$$\begin{gathered} -x^2+3x+40 \\ =-x^2+8x-5x+40 \\ =-x\left(x-8\right)-5\left(x-8\right) \\ =\left(x-8\right)\left(-x-5\right) \\ =\left(8-x\right)\left(x+5\right) \end{gathered}$$

Question 20:

Factorise:
x² – 26x + 133

Answer 20:

$$\begin{gathered} x^2-26x+133 \\ =x^2-19x-7x+133 \\ =x\left(x-19\right)-7\left(x-19\right) \\ =\left(x-19\right)\left(x-7\right) \end{gathered}$$

Question 21:

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

Answer 21:

$$\begin{gathered} x^2-2\sqrt{3}x-24 \\ =x^2-4\sqrt{3}x+2\sqrt{3}x-24 \\ =x\left(x-4\sqrt{3}\right)+2\sqrt{3}\left(x-4\sqrt{3}\right) \\ =\left(x-4\sqrt{3}\right)\left(x+2\sqrt{3}\right) \end{gathered}$$

Question 22:

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

Answer 22:

$$\begin{gathered} x^2-3\sqrt{5}x-20 \\ =x^2-4\sqrt{5}x+\sqrt{5}x-20 \\ =x\left(x-4\sqrt{5}\right)+\sqrt{5}\left(x-4\sqrt{5}\right) \\ =\left(x-4\sqrt{5}\right)\left(x+\sqrt{5}\right) \end{gathered}$$

Question 23:

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

Answer 23:

$$\begin{gathered} x^2+\sqrt{2}x-24 \\ =x^2+4\sqrt{2}x-3\sqrt{2}x-24 \\ =x\left(x+4\sqrt{2}\right)-3\sqrt{2}\left(x+4\sqrt{2}\right) \\ =\left(x+4\sqrt{2}\right)\left(x-3\sqrt{2}\right) \end{gathered}$$

Question 24:

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

Answer 24:

$$\begin{gathered} x^2-2\sqrt{2}x-30 \\ =x^2-5\sqrt{2}x+3\sqrt{2}x-30 \\ =x\left(x-5\sqrt{2}\right)+3\sqrt{2}\left(x-5\sqrt{2}\right) \\ =\left(x-5\sqrt{2}\right)\left(x+3\sqrt{2}\right) \end{gathered}$$

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, $-13+12=-1 \mathrm{and} -13\times 12=-156$.

$$\begin{gathered} \therefore x^2-x-156=x^2-13x+12x-156 \\ =x(x-13)+12(x-13) \\ =(x-13)(x+12) \end{gathered}$$

Question 26:

Factorise:
x² – 32x – 105

Answer 26:

$$\begin{gathered} x^2-32x-105 \\ =x^2-35x+3x-105 \\ =x\left(x-35\right)+3\left(x-35\right) \\ =\left(x-35\right)\left(x+3\right) \end{gathered}$$

Question 27:

Factorise:
9x² + 18x + 8

Answer 27:

$$\begin{gathered} 9x^2+18x+8 \\ =9x^2+12x+6x+8 \\ =3x\left(3x+4\right)+2\left(3x+4\right) \\ =\left(3x+4\right)\left(3x+2\right) \end{gathered}$$

Question 28:

Factorise:
6x² + 17x + 12

Answer 28:

$$\begin{gathered} 6x^2+17x+12 \\ =6x^2+9x+8x+12 \\ =3x\left(2x+3\right)+4\left(2x+3\right) \\ =\left(2x+3\right)\left(3x+4\right) \end{gathered}$$

Question 29:

Factorize:
18x² + 3x − 10

Answer 29:

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

​$$\begin{gathered} \therefore 18x^2+3x-10=18x^2+15x-12x-10 \\ =3x\left(6x+5\right)-2\left(6x+5\right) \\ =\left(6x+5\right)\left(3x-2\right) \end{gathered}$$

Question 30:

Factorize:
2x² + 11x − 21

Answer 30:

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

$$\begin{gathered} \therefore 2x^2+11x-21=2x^2+14x-3x-21 \\ =2x\left(x+7\right)-3\left(x+7\right) \\ =\left(x+7\right)\left(2x-3\right) \end{gathered}$$

Question 31:

Factorize:
15x² + 2x − 8

Answer 31:

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

$$\begin{gathered} \therefore 15x^2+2x-8=15x^2+12x-10x-8 \\ =3x\left(5x+4\right)-2\left(5x+4\right) \\ =\left(5x+4\right)\left(3x-2\right) \end{gathered}$$

Question 32:

Factorise:
21x² + 5x – 6

Answer 32:

$$\begin{gathered} 21x^2+5x-6 \\ =21x^2+14x-9x-6 \\ =7x\left(3x+2\right)-3\left(3x+2\right) \\ =\left(3x+2\right)\left(7x-3\right) \end{gathered}$$

Question 33:

Factorize:
24x² − 41x + 12

Answer 33:

We have:
$24x^2-41x+12$
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, $\left(-32\right)+\left(-9\right)=-41 \mathrm{and} \left(-32\right)\times \left(-9\right)=288$.

$$\begin{gathered} \therefore 24x^2-41x+12=24x^2-32x-9x+12 \\ =8x\left(3x-4\right)-3\left(3x-4\right) \\ =\left(3x-4\right)\left(8x-3\right) \end{gathered}$$

Question 34:

Factorise:
3x² – 14x + 8

Answer 34:

$$\begin{gathered} 3x^2-14x+8=3x^2-12x-2x+8 \\ =3x\left(x-4\right)-2\left(x-4\right) \\ =\left(x-4\right)\left(3x-2\right) \end{gathered}$$

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

Question 35:

Factorize:
2x² + 3x − 90

Answer 35:

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

$$\begin{gathered} \therefore 2x^2+3x-90=2x^2-12x+15x-90 \\ =2x\left(x-6\right)+15\left(x-6\right) \\ =\left(x-6\right)\left(2x+15\right) \end{gathered}$$

Question 36:

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

Answer 36:

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

$$\begin{gathered} \therefore \sqrt{5}{x}^2+2x-3\sqrt{5}=\sqrt{5}{x}^2+5x-3x-3\sqrt{5} \\ =\sqrt{5}x\left(x+\sqrt{5}\right)-3\left(x+\sqrt{5}\right) \\ =\left(x+\sqrt{5}\right)\left(\sqrt{5}x-3\right) \end{gathered}$$

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.,$2\sqrt{3}\times \left(-5\sqrt{3}\right)$.
Clearly, $6+\left(-5\right)=1 \mathrm{and} 6\times \left(-5\right)=-30$.

$$\begin{gathered} \therefore 2\sqrt{3}{x}^2+x-5\sqrt{3}=2\sqrt{3}{x}^2+6x-5x-5\sqrt{3} \\ =2\sqrt{3}x\left(x+\sqrt{3}\right)-5\left(x+\sqrt{3}\right) \\ =\left(x+\sqrt{3}\right)\left(2\sqrt{3}x-5\right) \end{gathered}$$

Question 38:

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

Answer 38:

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

Question 39:

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

Answer 39:

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

$$\begin{gathered} \therefore 6\sqrt{3}{x}^2-47x+5\sqrt{3} =6\sqrt{3}{x}^2-2x-45x+5\sqrt{3} \\ =2x\left(3\sqrt{3}x-1\right)-5\sqrt{3}\left(3\sqrt{3}x-1\right) \\ =\left(3\sqrt{3}x-1\right)\left(2x-5\sqrt{3}\right) \end{gathered}$$

Question 40:

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

Answer 40:

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

$$\begin{gathered} \therefore 5\sqrt{5}{x}^2+20x+3\sqrt{5}=5\sqrt{5}{x}^2+15x+5x+3\sqrt{5} \\ =5x\left(\sqrt{5}x+3\right)+\sqrt{5}(\sqrt{5}x+3) \\ =\left(\sqrt{5}x+3\right)\left(5x+\sqrt{5}\right) \end{gathered}$$

Question 41:

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

Answer 41:

$$\begin{gathered} \sqrt{3}{x}^2+10x+8\sqrt{3}=\sqrt{3}{x}^2+6x+4x+8\sqrt{3} \\ =\sqrt{3}x\left(x+2\sqrt{3}\right)+4\left(x+2\sqrt{3}\right) \\ =\left(x+2\sqrt{3}\right)\left(\sqrt{3}x+4\right) \end{gathered}$$

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

Question 42:

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

Answer 42:

We have:
$\sqrt{2}{x}^2+3x+\sqrt{2}$
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, $2+1=3 \mathrm{and} 2\times 1=2$.

$$\begin{gathered} \therefore \sqrt{2}{x}^2+3x+\sqrt{2}=\sqrt{2}{x}^2+2x+x+\sqrt{2} \\ =\sqrt{2}x\left(x+\sqrt{2}\right)+1\left(x+\sqrt{2}\right) \\ =\left(x+\sqrt{2}\right)\left(\sqrt{2}x+1\right) \end{gathered}$$

Question 43:

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

Answer 43:

We have:
$2x^2+3\sqrt{3}x+3$
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, $2\sqrt{3}+\sqrt{3}=3\sqrt{3} \mathrm{and} 2\sqrt{3}\times \sqrt{3}=6$.

$$\begin{gathered} \therefore 2x^2+3\sqrt{3}x+3=2x^2+2\sqrt{3}x+\sqrt{3}x+3 \\ =2x\left(x+\sqrt{3}\right)+\sqrt{3}\left(x+\sqrt{3}\right) \\ =\left(x+\sqrt{3}\right)\left(2x+\sqrt{3}\right) \end{gathered}$$

Question 44:

Factorize:
15x² − x − 128

Answer 44:

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

$$\begin{gathered} \therefore 15x^2-x-28=15x^2-21x+20x-28 \\ =3x(5x-7)+4(5x-7) \\ =(5x-7)(3x+4) \end{gathered}$$

Question 45:

Factorize:
6x² − 5x − 21

Answer 45:

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

$$\begin{gathered} \therefore 6x^2-5x-21=6x^2+9x-14x-21 \\ =3x\left(2x+3\right)-7\left(2x+3\right) \\ =\left(2x+3\right)\left(3x-7\right) \end{gathered}$$

Question 46:

Factorize:
2x² − 7x − 15

Answer 46:

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

$$\begin{gathered} \therefore 2x^2-7x-15=2x^2-10x+3x-15 \\ =2x\left(x-5\right)+3\left(x-5\right) \\ =\left(x-5\right)\left(2x+3\right) \end{gathered}$$

Question 47:

Factorize:
5x² − 16x − 21

Answer 47:

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

$$\begin{gathered} \therefore 5x^2-16x-21=5x^2+5x-21x-21 \\ =5x\left(x+1\right)-21\left(x+1\right) \\ =\left(x+1\right)\left(5x-21\right) \end{gathered}$$

Question 48:

Factorise:
6x² – 11x – 35

Answer 48:

$$\begin{gathered} 6x^2-11x-35=6x^2-21x+10x-35 \\ =3x\left(2x-7\right)+5\left(2x-7\right) \\ =\left(2x-7\right)\left(3x+5\right) \end{gathered}$$

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

Question 49:

Factorise:
9x² – 3x – 20

Answer 49:

$$\begin{gathered} 9x^2-3x-20=9x^2-15x+12x-20 \\ =3x\left(3x-5\right)+4\left(3x-5\right) \\ =\left(3x-5\right)\left(3x+4\right) \end{gathered}$$

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

Question 50:

Factorize:
10x² − 9x − 7

Answer 50:

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

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

$$\begin{gathered} \therefore 10x^2-9x-7=10x^2+5x-14x-7 \\ =5x\left(2x+1\right)-7\left(2x+1\right) \\ =\left(2x+1\right)\left(5x-7\right) \end{gathered}$$

Question 51:

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

Answer 51:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {x}^2-2x+\frac{7}{16} \\ =\frac{16x^2-32x+7}{16} \\ =\frac{1}{16}\left(16x^2-32x+7\right) \end{gathered}$$

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, $\left(-4\right)+\left(-28\right)=-32 \mathrm{and} \left(-4\right)\times \left(-28\right)=112$.

$$\begin{gathered} \therefore x^2 - 2x + \frac{7}{16} =\frac{1}{16}(16x^2-32x+7) \\ =\frac{1}{16}(16x^2-4x-28x+7) \\ =\frac{1}{16}\left[4x(4x-1)-7(4x-1)\right] \\ =\frac{1}{16}(4x-1)(4x-7) \end{gathered}$$

Question 52:

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

Answer 52:

$$\begin{gathered} \frac{1}{3}{x}^2-2x-9=\frac{x^2-6x-27}{3} \\ =\frac{x^2-9x+3x-27}{3} \\ =\frac{x\left(x-9\right)+3\left(x-9\right)}{3} \\ =\frac{\left(x-9\right)\left(x+3\right)}{3} \\ =\frac{\left(x-9\right)}{3}\times \frac{\left(x+3\right)}{1} \\ =\left(\frac{1}{3}x-3\right)\left(x+3\right) \end{gathered}$$

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

Question 53:

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

Answer 53:

$$\begin{gathered} x^2+\frac{12}{35}x+\frac{1}{35}=\frac{35x^2+12x+1}{35} \\ =\frac{35x^2+7x+5x+1}{35} \\ =\frac{7x\left(5x+1\right)+1\left(5x+1\right)}{35} \\ =\frac{\left(5x+1\right)\left(7x+1\right)}{35} \\ =\frac{\left(5x+1\right)\left(7x+1\right)}{5\times 7} \\ =\frac{\left(5x+1\right)}{5}\times \frac{\left(7x+1\right)}{7} \\ =\left(x+\frac{1}{5}\right)\left(x+\frac{1}{7}\right) \end{gathered}$$

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

Question 54:

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

Answer 54:

$$\begin{gathered} 21x^2-2x+\frac{1}{21}=21x^2-x-x+\frac{1}{21} \\ =21x\left(x-\frac{1}{21}\right)-1\left(x-\frac{1}{21}\right) \\ =\left(x-\frac{1}{21}\right)\left(21x-1\right) \end{gathered}$$

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

Question 55:

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

Answer 55:

$$\begin{gathered} \frac{3}{2}{x}^2+16x+10=\frac{3}{2}{x}^2+15x+x+10 \\ =3x\left(\frac{1}{2}x+5\right)+1\left(x+10\right) \\ =\frac{3}{2}x\left(x+10\right)+1\left(x+10\right) \\ =\left(x+10\right)\left(\frac{3}{2}x+1\right) \end{gathered}$$

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

Question 56:

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

Answer 56:

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

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

Question 57:

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

Answer 57:

$$\begin{gathered} \frac{3}{5}{x}^2-\frac{19}{5}x+4=\frac{3}{5}{x}^2-3x-\frac{4}{5}x+4 \\ =3x\left(\frac{1}{5}x-1\right)-4\left(\frac{1}{5}x-1\right) \\ =\left(\frac{1}{5}x-1\right)\left(3x-4\right) \end{gathered}$$

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

Question 58:

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

Answer 58:

$$\begin{gathered} 2x^2-x+\frac{1}{8}=2x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{8} \\ =2x\left(x-\frac{1}{4}\right)-\frac{1}{2}\left(x-\frac{1}{4}\right) \\ =\left(x-\frac{1}{4}\right)\left(2x-\frac{1}{2}\right) \end{gathered}$$

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

Question 59:

Factorize:
2(x + y)² − 9(x + y) − 5

Answer 59:

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

$$\begin{gathered} \therefore 2u^2-9u-5=2u^2-10u+u-5 \\ =2u(u-5)+1(u-5) \\ =(u-5)(2u+1) \end{gathered}$$
Putting $u=(x+y)$, we get:
$$\begin{gathered} 2{\left(x+y\right)}^2 - 9\left(x+y\right) - 5 = \left(x+y-5\right)\left[2\left(x+y\right)+1\right] \\ = \left(x+y-5\right)\left(2x+2y+1\right) \end{gathered}$$

Question 60:

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

Answer 60:

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

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Question 61:

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

Answer 61:

$$\begin{gathered} 7{\left(x-2y\right)}^2-25\left(x-2y\right)+12=7{\left(x-2y\right)}^2-21\left(x-2y\right)-4\left(x-2y\right)+12 \\ =\left[7\left(x-2y\right)\right]\left(x-2y-3\right)-4\left(x-2y-3\right) \\ =\left[7\left(x-2y\right)-4\right]\left(x-2y-3\right) \\ =\left(7x-14y-4\right)\left(x-2y-3\right) \end{gathered}$$

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

Question 62:

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

Answer 62:

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

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

Question 63:

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

Answer 63:

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

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

Question 64:

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

Answer 64:

$$\begin{gathered} {\left(a+2b\right)}^2+101\left(a+2b\right)+100={\left(a+2b\right)}^2+100\left(a+2b\right)+1\left(a+2b\right)+100 \\ =\left(a+2b\right)\left[\left(a+2b\right)+100\right]+1\left[\left(a+2b\right)+100\right] \\ =\left[\left(a+2b\right)+1\right]\left[\left(a+2b\right)+100\right] \\ =\left(a+2b+1\right)\left(a+2b+100\right) \end{gathered}$$

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

Question 65:

Factorise:
4x⁴ + 7x² – 2

Answer 65:

$$\begin{gathered} 4x^4+7x^2-2=4x^4+8x^2-x^2-2 \\ =4x^2\left(x^2+2\right)-1\left(x^2+2\right) \\ =\left(4x^2-1\right)\left(x^2+2\right) \end{gathered}$$

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

Question 66:

Evaluate {(999)2 – 1}.

Answer 66:

$$\begin{gathered} \left\{{\left(999\right)}^2-1\right\}=\left\{{\left(999\right)}^2-1^2\right\} \\ =\left(999-1\right)\left(999+1\right) \\ =\left(998\right)\left(1000\right) \\ =998000 \end{gathered}$$

Hence, {(999)² – 1} = 998000.