Rs Aggarwal 2019 2020 Solutions for Class 9 Math Chapter 3 Factorisation Of Polynomials are provided here with simple step-by-step explanations. These solutions for Factorisation Of Polynomials are extremely popular among Class 9 students for Math Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 9 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Question 1:

Factorize:
9x2 + 12xy

We have:
$9{x}^{2}+12xy\phantom{\rule{0ex}{0ex}}=3x\left(3x+4y\right)$

#### Question 2:

Factorize:
18x2y − 24xyz

We have:
$18{x}^{2}y-24xyz\phantom{\rule{0ex}{0ex}}=6xy\left(3y-4z\right)$

#### Question 3:

Factorize:
27a3b3 − 45a4b2

We have:
$27{a}^{3}{b}^{3}-45{a}^{4}{b}^{2}\phantom{\rule{0ex}{0ex}}=9{a}^{3}{b}^{2}\left(3b-5a\right)$

#### Question 4:

Factorize:
2a(x + y) − 3b(x + y)

We have:
$2a\left(x+y\right)-3b\left(x+y\right)\phantom{\rule{0ex}{0ex}}=\left(x+y\right)\left(2a-3b\right)$

#### Question 5:

Factorize:
2x(p2 + q2) + 4y(p2 + q2)

We have:
$2x\left({p}^{2}+{q}^{2}\right)+4y\left({p}^{2}+{q}^{2}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left[x\left({p}^{2}+{q}^{2}\right)+2y\left({p}^{2}+{q}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=2\left({p}^{2}+{q}^{2}\right)\left(x+2y\right)\phantom{\rule{0ex}{0ex}}$

#### Question 6:

Factorize:
x(a − 5) + y(5 − a)

We have:
$x\left(a-5\right)+y\left(5-a\right)=x\left(a-5\right)-y\left(a-5\right)$
$=\left(a-5\right)\left(x-y\right)$

#### Question 7:

Factorize:
4(a + b) − 6(a + b)2

We have:
$4\left(a+b\right)-6{\left(a+b\right)}^{2}=2\left(a+b\right)\left[2-3\left(a+b\right)\right]$
$=2\left(a+b\right)\left(2-3a-3b\right)$

#### Question 8:

Factorize:
8(3a − 2b)2 − 10(3a − 2b)

We have:
$8{\left(3a-2b\right)}^{2}-10\left(3a-2b\right)=2\left(3a-2b\right)\left[4\left(3a-2b\right)-5\right]$
$=2\left(3a-2b\right)\left(12a-8b-5\right)$

#### Question 9:

Factorize:
x(x + y)3 − 3x2y(x + y)

We have:
$x{\left(x+y\right)}^{3}-3{x}^{2}y\left(x+y\right)=x\left(x+y\right)\left[{\left(x+y\right)}^{2}-3xy\right]\phantom{\rule{0ex}{0ex}}$
$=x\left(x+y\right)\left[{x}^{2}+{y}^{2}+2xy-3xy\right]\phantom{\rule{0ex}{0ex}}=x\left(x+y\right)\left({x}^{2}+{y}^{2}-xy\right)$

#### Question 10:

Factorize:
x3 + 2x2 + 5x + 10

We have:
${x}^{3}+2{x}^{2}+5x+10=\left({x}^{3}+2{x}^{2}\right)+\left(5x+10\right)$
$={x}^{2}\left(x+2\right)+5\left(x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x+2\right)\left({x}^{2}+5\right)$

#### Question 11:

Factorize:
x2 + xy − 2xz − 2yz

We have:

#### Question 12:

Factorize:
a3ba2b + 5ab − 5b

We have:

$=b\left[{a}^{2}\left(a-1\right)+5\left(a-1\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=b\left(a-1\right)\left({a}^{2}+5\right)$

#### Question 13:

Factorize:
8 − 4a − 2a3 + a4

We have:

#### Question 14:

Factorize:
x3 − 2x2y + 3xy2 − 6y3

We have:
${x}^{3}-2{x}^{2}y+3x{y}^{2}-6{y}^{3}=\left({x}^{3}-2{x}^{2}y\right)+\left(3x{y}^{2}-6{y}^{3}\right)$
$={x}^{2}\left(x-2y\right)+3{y}^{2}\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left({x}^{2}+3{y}^{2}\right)$

#### Question 15:

Factorize:
px − 5q + pq − 5x

We have:
$px-5q+pq-5x=\left(px-5x\right)+\left(pq-5q\right)$
$=x\left(p-5\right)+q\left(p-5\right)\phantom{\rule{0ex}{0ex}}=\left(p-5\right)\left(x+q\right)$

#### Question 16:

Factorize:
x2 + yxyx

We have:
${x}^{2}+y-xy-x=\left({x}^{2}-xy\right)-\left(x-y\right)$
$=x\left(x-y\right)-1\left(x-y\right)\phantom{\rule{0ex}{0ex}}=\left(x-y\right)\left(x-1\right)$

#### Question 17:

Factorize:
(3a − 1)2 − 6a + 2

We have:
${\left(3a-1\right)}^{2}-6a+2={\left(3a-1\right)}^{2}-2\left(3a-1\right)$
$=\left(3a-1\right)\left[\left(3a-1\right)-2\right]\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-1-2\right)\phantom{\rule{0ex}{0ex}}=\left(3a-1\right)\left(3a-3\right)\phantom{\rule{0ex}{0ex}}=3\left(3a-1\right)\left(a-1\right)$

#### Question 18:

Factorize:
(2x − 3)2 − 8x + 12

We have:
${\left(2x-3\right)}^{2}-8x+12={\left(2x-3\right)}^{2}-4\left(2x-3\right)$
$=\left(2x-3\right)\left[\left(2x-3\right)-4\right]\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-3-4\right)\phantom{\rule{0ex}{0ex}}=\left(2x-3\right)\left(2x-7\right)$

#### Question 19:

Factorize:
a3 + a − 3a2 − 3

We have:
${a}^{3}+a-3{a}^{2}-3=\left({a}^{3}-3{a}^{2}\right)+\left(a-3\right)\phantom{\rule{0ex}{0ex}}$
$={a}^{2}\left(a-3\right)+1\left(a-3\right)\phantom{\rule{0ex}{0ex}}=\left(a-3\right)\left({a}^{2}+1\right)$

#### Question 20:

Factorize:
3ax − 6ay − 8by + 4bx

We have:
$3ax-6ay-8by+4bx=\left(3ax-6ay\right)+\left(4bx-8by\right)$
$=3a\left(x-2y\right)+4b\left(x-2y\right)\phantom{\rule{0ex}{0ex}}=\left(x-2y\right)\left(3a+4b\right)$

#### Question 21:

Factorize:
abx2 + a2x + b2x + ab

We have:
$ab{x}^{2}+{a}^{2}x+{b}^{2}x+ab=\left(ab{x}^{2}+{b}^{2}x\right)+\left({a}^{2}x+ab\right)$
$=bx\left(ax+b\right)+a\left(ax+b\right)\phantom{\rule{0ex}{0ex}}=\left(ax+b\right)\left(bx+a\right)$

#### Question 22:

Factorize:
x3x2 + ax + xa − 1

We have:
${x}^{3}-{x}^{2}+ax+x-a-1=\left({x}^{3}-{x}^{2}\right)+\left(ax-a\right)+\left(x-1\right)$
$={x}^{2}\left(x-1\right)+a\left(x-1\right)+1\left(x-1\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left({x}^{2}+a+1\right)$

#### Question 23:

Factorize:
2x + 4y − 8xy − 1

We have:
$2x+4y-8xy-1=\left(2x-8xy\right)-\left(1-4y\right)$
$=2x\left(1-4y\right)-1\left(1-4y\right)\phantom{\rule{0ex}{0ex}}=\left(1-4y\right)\left(2x-1\right)$

#### Question 24:

Factorize:
ab(x2 + y2) − xy(a2 + b2)

We have:
$ab\left({x}^{2}+{y}^{2}\right)-xy\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab{y}^{2}-{a}^{2}xy-{b}^{2}xy$
$=\left(ab{x}^{2}-{a}^{2}xy\right)-\left({b}^{2}xy-ab{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx-ay\right)-by\left(bx-ay\right)\phantom{\rule{0ex}{0ex}}=\left(bx-ay\right)\left(ax-by\right)$

#### Question 25:

Factorize:
a2 + ab(b + 1) + b3

We have:
${a}^{2}+ab\left(b+1\right)+{b}^{3}={a}^{2}+a{b}^{2}+ab+{b}^{3}$
$=\left({a}^{2}+a{b}^{2}\right)+\left(ab+{b}^{3}\right)\phantom{\rule{0ex}{0ex}}=a\left(a+{b}^{2}\right)+b\left(a+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(a+{b}^{2}\right)\left(a+b\right)$

#### Question 26:

Factorize:
a3 + ab(1 − 2a) − 2b2

We have:
${a}^{3}+ab\left(1-2a\right)-2{b}^{2}={a}^{3}+ab-2{a}^{2}b-2{b}^{2}$
$=\left({a}^{3}-2{a}^{2}b\right)+\left(ab-2{b}^{2}\right)\phantom{\rule{0ex}{0ex}}={a}^{2}\left(a-2b\right)+b\left(a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(a-2b\right)\left({a}^{2}+b\right)$

#### Question 27:

Factorize:
2a2 + bc − 2abac2

We have:
$2{a}^{2}+bc-2ab-ac=\left(2{a}^{2}-2ab\right)-\left(ac-bc\right)$
$=2a\left(a-b\right)-c\left(a-b\right)\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\left(2a-c\right)$

#### Question 28:

Factorize:
(ax + by)2 + (bxay)2

We have:
${\left(ax+by\right)}^{2}+{\left(bx-ay\right)}^{2}=\left[{\left(ax\right)}^{2}+2×ax×by+{\left(by\right)}^{2}\right]+\left[{\left(bx\right)}^{2}-2×bx×ay+{\left(ay\right)}^{2}\right]$
$={a}^{2}{x}^{2}+2abxy+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}-2abxy+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}={a}^{2}{x}^{2}+{b}^{2}{y}^{2}+{b}^{2}{x}^{2}+{a}^{2}{y}^{2}\phantom{\rule{0ex}{0ex}}=\left({a}^{2}{x}^{2}+{b}^{2}{x}^{2}\right)+\left({a}^{2}{y}^{2}+{b}^{2}{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({a}^{2}+{b}^{2}\right)+{y}^{2}\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left({a}^{2}+{b}^{2}\right)\left({x}^{2}+{y}^{2}\right)$

#### Question 29:

Factorize:
a(a + bc) − bc

We have:
$a\left(a+b-c\right)-bc={a}^{2}+ab-ac-bc$
$=\left({a}^{2}-ac\right)+\left(ab-bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-c\right)+b\left(a-c\right)\phantom{\rule{0ex}{0ex}}=\left(a-c\right)\left(a+b\right)$

#### Question 30:

Factorize:
a(a − 2bc) + 2bc

We have:
$a\left(a-2b-c\right)+2bc={a}^{2}-2ab-ac+2bc$
$=\left({a}^{2}-2ab\right)-\left(ac-2bc\right)\phantom{\rule{0ex}{0ex}}=a\left(a-2b\right)-c\left(a-2b\right)\phantom{\rule{0ex}{0ex}}=\left(a-2b\right)\left(a-c\right)$

#### Question 31:

Factorize:
a2x2 + (ax2 + 1)x + a

We have:
${a}^{2}{x}^{2}+\left(a{x}^{2}+1\right)x+a=\left(a{x}^{2}+1\right)x+\left({a}^{2}{x}^{2}+a\right)$
$=x\left(a{x}^{2}+1\right)+a\left(a{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=\left(a{x}^{2}+1\right)\left(x+a\right)$

#### Question 32:

Factorize:
ab(x2 + 1) + x(a2 + b2)

We have:
$ab\left({x}^{2}+1\right)+x\left({a}^{2}+{b}^{2}\right)=ab{x}^{2}+ab+{a}^{2}x+{b}^{2}x$
$=\left(ab{x}^{2}+{a}^{2}x\right)+\left({b}^{2}x+ab\right)\phantom{\rule{0ex}{0ex}}=ax\left(bx+a\right)+b\left(bx+a\right)\phantom{\rule{0ex}{0ex}}=\left(bx+a\right)\left(ax+b\right)$

#### Question 33:

Factorize:
x2 − (a + b)x + ab

We have:
${x}^{2}-\left(a+b\right)x+ab={x}^{2}-ax-bx+ab$
$=\left({x}^{2}-ax\right)-\left(bx-ab\right)\phantom{\rule{0ex}{0ex}}=x\left(x-a\right)-b\left(x-a\right)\phantom{\rule{0ex}{0ex}}=\left(x-a\right)\left(x-b\right)$

#### Question 34:

Factorize:
${x}^{2}+\frac{1}{{x}^{2}}-2-3x+\frac{3}{x}$

Factorise:
9x2 – 16y2

#### Question 2:

Factorise:
$\left(\frac{25}{4}{x}^{2}-\frac{1}{9}{y}^{2}\right)$

Factorise:
81 – 16x2

Factorise:
5 – 20x2

Factorise:
2x4 – 32

Factorize:
3a3b − 243ab3

Factorize:
3x3 − 48x

Factorize:
27a2 − 48b2

Factorize:
x − 64x3

Factorize:
8ab2 − 18a3

Factorize:
150 − 6x2

Factorise:
2 – 50x2

Factorise:
20x2 – 45

Factorise:
(3a + 5b)2 – 4c2

Factorise:
a2 b2 a – b

#### Question 16:

Factorise:
4a2 – 9b2 – 2a – 3b

Factorise:
a2 b2 + 2bcc2

#### Question 18:

Factorise:
4a2 – 4b2 + 4a + 1

#### Question 19:

Factorize:
a2 + 2ab + b2 − 9c2

Factorize:
108a2 − 3(bc)2

Factorize:
(a + b)3ab

Factorise:
x2 + y2z2 – 2xy

#### Question 23:

Factorise:
x2 + 2xy + y2 a2 + 2abb2

#### Question 24:

Factorise:
25x2 – 10x + 1 – 36y2

Factorize:
aba2 + b2

Factorize:
a2b2 − 4ac + 4c2

Factorize:
9 − a2 + 2abb2

Factorize:
x3 − 5x2x + 5

#### Question 29:

Factorise:
1 + 2ab – (a2 + b2)

$1+2ab-\left({a}^{2}+{b}^{2}\right)\phantom{\rule{0ex}{0ex}}=1+2ab-{a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}=1-{a}^{2}+2ab-{b}^{2}\phantom{\rule{0ex}{0ex}}={1}^{2}-\left({a}^{2}-2ab+{b}^{2}\right)$

#### Question 30:

Factorise:
9a2 + 6a + 1 – 36b2

Factorize:
x2y2 + 6y − 9

#### Question 32:

Factorize:
4x2 − 9y2 − 2x − 3y

#### Question 33:

Factorize:
9a2 + 3a − 8b − 64b2

#### Question 34:

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

#### Question 35:

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

Disclaimer: The expression of the question should be ${x}^{2}-2+\frac{1}{{x}^{2}}-{y}^{2}$. The same has been done before solving the question.

#### Question 36:

Factorise:
${x}^{4}+\frac{4}{{x}^{4}}$

Factorise:
x8 – 1

Factorise:
16x4 – 1

81x4y4

x4 – 625

#### Question 1:

Factorize:
x2 + 11x + 30

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, .

#### Question 2:

Factorize:
x2 + 18x + 32

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, .

#### Question 3:

Factorise:
x
2 + 20x – 69

${x}^{2}+20x-69\phantom{\rule{0ex}{0ex}}={x}^{2}+23x-3x-69\phantom{\rule{0ex}{0ex}}=x\left(x+23\right)-3\left(x+23\right)\phantom{\rule{0ex}{0ex}}=\left(x+23\right)\left(x-3\right)$

#### Question 4:

x2 + 19x – 150

${x}^{2}+19x-150\phantom{\rule{0ex}{0ex}}={x}^{2}+25x-6x-150\phantom{\rule{0ex}{0ex}}=x\left(x+25\right)-6\left(x+25\right)\phantom{\rule{0ex}{0ex}}=\left(x+25\right)\left(x-6\right)$

#### Question 5:

Factorise:
x
2 + 7x – 98

${x}^{2}+7x-98\phantom{\rule{0ex}{0ex}}={x}^{2}+14x-7x-98\phantom{\rule{0ex}{0ex}}=x\left(x+14\right)-7\left(x+14\right)\phantom{\rule{0ex}{0ex}}=\left(x+14\right)\left(x-7\right)$

#### Question 6:

Factorise:
${x}^{2}+2\sqrt{3}x–24$

${x}^{2}+2\sqrt{3}x–24\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{3}x-2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{3}\right)-2\sqrt{3}\left(x+4\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{3}\right)\left(x-2\sqrt{3}\right)$

#### Question 7:

Factorise:
x
2 21x + 90

${x}^{2}-21x+90\phantom{\rule{0ex}{0ex}}={x}^{2}-15x-6x+90\phantom{\rule{0ex}{0ex}}=x\left(x-15\right)-6\left(x-15\right)\phantom{\rule{0ex}{0ex}}=\left(x-6\right)\left(x-15\right)$

#### Question 8:

Factorise:
x
2 – 22x + 120

${x}^{2}-22x+120\phantom{\rule{0ex}{0ex}}={x}^{2}-12x-10x+120\phantom{\rule{0ex}{0ex}}=x\left(x-12\right)-10\left(x-12\right)\phantom{\rule{0ex}{0ex}}=\left(x-10\right)\left(x-12\right)$

#### Question 9:

Factorise:
x
2 4x + 3

${x}^{2}-4x+3\phantom{\rule{0ex}{0ex}}={x}^{2}-3x-x+3\phantom{\rule{0ex}{0ex}}=x\left(x-3\right)-1\left(x-3\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-3\right)$

#### Question 10:

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

${x}^{2}+7\sqrt{6}x+60\phantom{\rule{0ex}{0ex}}={x}^{2}+5\sqrt{6}x+2\sqrt{6}x+60\phantom{\rule{0ex}{0ex}}=x\left(x+5\sqrt{6}\right)+2\sqrt{6}\left(x+5\sqrt{6}\right)\phantom{\rule{0ex}{0ex}}=\left(x+5\sqrt{6}\right)\left(x+2\sqrt{6}\right)$

#### Question 11:

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

${x}^{2}+3\sqrt{3}x+6\phantom{\rule{0ex}{0ex}}={x}^{2}+2\sqrt{3}x+\sqrt{3}x+6\phantom{\rule{0ex}{0ex}}=x\left(x+2\sqrt{3}\right)+\sqrt{3}\left(x+2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x+2\sqrt{3}\right)\left(x+\sqrt{3}\right)$

#### Question 12:

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

${x}^{2}+6\sqrt{6}x+48\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{6}x+2\sqrt{6}x+48\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{6}\right)+2\sqrt{6}\left(x+4\sqrt{6}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{6}\right)\left(x+2\sqrt{6}\right)$

#### Question 13:

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

${x}^{2}+5\sqrt{5}x+30\phantom{\rule{0ex}{0ex}}={x}^{2}+3\sqrt{5}x+2\sqrt{5}x+30\phantom{\rule{0ex}{0ex}}=x\left(x+3\sqrt{5}\right)+2\sqrt{5}\left(x+3\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\sqrt{5}\right)\left(x+2\sqrt{5}\right)$

#### Question 14:

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

${x}^{2}-24x-180\phantom{\rule{0ex}{0ex}}={x}^{2}-30x+6x-180\phantom{\rule{0ex}{0ex}}=x\left(x-30\right)+6\left(x-30\right)\phantom{\rule{0ex}{0ex}}=\left(x-30\right)\left(x+6\right)$

#### Question 15:

Factorise:
x
2 – 32x – 105

${x}^{2}-32x-105\phantom{\rule{0ex}{0ex}}={x}^{2}-35x+3x-105\phantom{\rule{0ex}{0ex}}=x\left(x-35\right)+3\left(x-35\right)\phantom{\rule{0ex}{0ex}}=\left(x-35\right)\left(x+3\right)$

#### Question 16:

Factorise:
x
2 – 11x – 80

${x}^{2}-11x-80\phantom{\rule{0ex}{0ex}}={x}^{2}-16x+5x-80\phantom{\rule{0ex}{0ex}}=x\left(x-16\right)+5\left(x-16\right)\phantom{\rule{0ex}{0ex}}=\left(x-16\right)\left(x+5\right)$

#### Question 17:

Factorise:
6 – x – x2

$-{x}^{2}-x+6\phantom{\rule{0ex}{0ex}}=-{x}^{2}-3x+2x+6\phantom{\rule{0ex}{0ex}}=-x\left(x+3\right)+2\left(x+3\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\right)\left(-x+2\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\right)\left(2-x\right)$

#### Question 18:

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

${x}^{2}-\sqrt{3}x-6\phantom{\rule{0ex}{0ex}}={x}^{2}-2\sqrt{3}x+\sqrt{3}x-6\phantom{\rule{0ex}{0ex}}=x\left(x-2\sqrt{3}\right)+\sqrt{3}\left(x-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x-2\sqrt{3}\right)\left(x+\sqrt{3}\right)$

#### Question 19:

Factorise:
403x – x2

$-{x}^{2}+3x+40\phantom{\rule{0ex}{0ex}}=-{x}^{2}+8x-5x+40\phantom{\rule{0ex}{0ex}}=-x\left(x-8\right)-5\left(x-8\right)\phantom{\rule{0ex}{0ex}}=\left(x-8\right)\left(-x-5\right)\phantom{\rule{0ex}{0ex}}=\left(8-x\right)\left(x+5\right)$

#### Question 20:

Factorise:
x226x + 133

${x}^{2}-26x+133\phantom{\rule{0ex}{0ex}}={x}^{2}-19x-7x+133\phantom{\rule{0ex}{0ex}}=x\left(x-19\right)-7\left(x-19\right)\phantom{\rule{0ex}{0ex}}=\left(x-19\right)\left(x-7\right)$

#### Question 21:

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

${x}^{2}-2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}={x}^{2}-4\sqrt{3}x+2\sqrt{3}x-24\phantom{\rule{0ex}{0ex}}=x\left(x-4\sqrt{3}\right)+2\sqrt{3}\left(x-4\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=\left(x-4\sqrt{3}\right)\left(x+2\sqrt{3}\right)$

#### Question 22:

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

${x}^{2}-3\sqrt{5}x-20\phantom{\rule{0ex}{0ex}}={x}^{2}-4\sqrt{5}x+\sqrt{5}x-20\phantom{\rule{0ex}{0ex}}=x\left(x-4\sqrt{5}\right)+\sqrt{5}\left(x-4\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=\left(x-4\sqrt{5}\right)\left(x+\sqrt{5}\right)$

#### Question 23:

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

${x}^{2}+\sqrt{2}x-24\phantom{\rule{0ex}{0ex}}={x}^{2}+4\sqrt{2}x-3\sqrt{2}x-24\phantom{\rule{0ex}{0ex}}=x\left(x+4\sqrt{2}\right)-3\sqrt{2}\left(x+4\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}=\left(x+4\sqrt{2}\right)\left(x-3\sqrt{2}\right)$

#### Question 24:

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

${x}^{2}-2\sqrt{2}x-30\phantom{\rule{0ex}{0ex}}={x}^{2}-5\sqrt{2}x+3\sqrt{2}x-30\phantom{\rule{0ex}{0ex}}=x\left(x-5\sqrt{2}\right)+3\sqrt{2}\left(x-5\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}=\left(x-5\sqrt{2}\right)\left(x+3\sqrt{2}\right)$

#### Question 25:

Factorize:
x2x − 156

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, .

#### Question 26:

Factorise:
x2 – 32x – 105

${x}^{2}-32x-105\phantom{\rule{0ex}{0ex}}={x}^{2}-35x+3x-105\phantom{\rule{0ex}{0ex}}=x\left(x-35\right)+3\left(x-35\right)\phantom{\rule{0ex}{0ex}}=\left(x-35\right)\left(x+3\right)$

#### Question 27:

Factorise:
9x2 + 18x + 8

$9{x}^{2}+18x+8\phantom{\rule{0ex}{0ex}}=9{x}^{2}+12x+6x+8\phantom{\rule{0ex}{0ex}}=3x\left(3x+4\right)+2\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=\left(3x+4\right)\left(3x+2\right)$

#### Question 28:

Factorise:
6x2 + 17x + 12

$6{x}^{2}+17x+12\phantom{\rule{0ex}{0ex}}=6{x}^{2}+9x+8x+12\phantom{\rule{0ex}{0ex}}=3x\left(2x+3\right)+4\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=\left(2x+3\right)\left(3x+4\right)$

#### Question 29:

Factorize:
18x2 + 3x − 10

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

#### Question 30:

Factorize:
2x2 + 11x − 21

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

#### Question 31:

Factorize:
15x2 + 2x − 8

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

#### Question 32:

Factorise:
21x2 + 5x – 6

$21{x}^{2}+5x-6\phantom{\rule{0ex}{0ex}}=21{x}^{2}+14x-9x-6\phantom{\rule{0ex}{0ex}}=7x\left(3x+2\right)-3\left(3x+2\right)\phantom{\rule{0ex}{0ex}}=\left(3x+2\right)\left(7x-3\right)$

#### Question 33:

Factorize:
24x2 − 41x + 12

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

#### Question 34:

Factorise:
3x2 – 14x + 8

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

#### Question 35:

Factorize:
2x2 + 3x − 90

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

#### Question 36:

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

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}×\left(-3\sqrt{5}\right)$.
Clearly, .

#### Question 37:

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

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.,$2\sqrt{3}×\left(-5\sqrt{3}\right)$.
Clearly, .

#### Question 38:

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

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, .

#### Question 39:

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

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, .

#### Question 40:

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

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,

#### Question 41:

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

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

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., $\sqrt{2}×\sqrt{2}$.
Clearly, .

#### Question 43:

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

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.,$2×3$.
Clearly, .

#### Question 44:

Factorize:
15x2x − 128

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., $15×\left(-28\right)$.
Clearly, .

#### Question 45:

Factorize:
6x2 − 5x − 21

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

#### Question 46:

Factorize:
2x2 − 7x − 15

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

#### Question 47:

Factorize:
5x2 − 16x − 21

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

#### Question 48:

Factorise:
6x2 – 11x – 35

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

#### Question 49:

Factorise:
9x2 – 3x – 20

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

#### Question 50:

Factorize:
10x2 − 9x − 7

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

We have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$70), i.e., $10×\left(-7\right)$.
Clearly, .

#### Question 51:

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

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

#### Question 52:

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

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

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

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

#### Question 55:

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

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

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

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

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

#### Question 59:

Factorize:
2(x + y)2 − 9(x + y) − 5

We have:
$2{\left(x+y\right)}^{2}-9\left(x+y\right)-5\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(x+y\right)=u\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$2{u}^{2}-9u-5\phantom{\rule{0ex}{0ex}}$
Now, we have to split ($-$9) into two numbers such that their sum is ($-$9) and their product is ($-$10).
Clearly, .

Putting $u=\left(x+y\right)$, we get:

#### Question 60:

Factorize:
9(2ab)2 − 4(2ab) − 13

We have:
$9\left(2a-b{\right)}^{2}-4\left(2a-b\right)-13\phantom{\rule{0ex}{0ex}}\mathrm{Let}:\phantom{\rule{0ex}{0ex}}\left(2a-b\right)=p\phantom{\rule{0ex}{0ex}}$
Thus, the given expression becomes
$9{p}^{2}-4p-13\phantom{\rule{0ex}{0ex}}$
Now, we must split ($-$4) into two numbers such that their sum is ($-$4) and their product is ($-$117).
Clearly, .

Putting $p=\left(2a-b\right)$, we get:

#### Question 61:

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

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

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

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

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

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

#### Question 66:

Evaluate {(999)2 – 1}.

Hence, {(999)2 – 1} = 998000.

#### Question 1:

Expand:
(i) (a + 2b + 5c)2
(ii) (2bb + c)2
(iii) (a − 2b − 3c)2

#### Question 2:

Expand:
(i) (2a − 5b − 7c)2
(ii) (−3a + 4b − 5c)2
(iii) ${\left(\frac{1}{2}a-\frac{1}{4}a+2\right)}^{2}$

#### Question 3:

Factorize: 4x2 + 9y2 + 16z2 + 12xy − 24yz − 16xz.

#### Question 4:

Factorize: 9x2 + 16y2 + 4z2 − 24xy + 16yz − 12xz

#### Question 5:

Factorize: 25x2 + 4y2 + 9z2 − 20xy − 12yz + 30xz.

#### Question 6:

16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz

Hence, 16x2 + 4y2 + 9z2 – 16xy – 12yz + 24xz = ${\left(4x-2y+3z\right)}^{2}$.

Evaluate
(i) (99)2
(ii) (995)2
(iii) (107)2

#### Question 1:

Expand
(i) (3x + 2)3
(ii) ${\left(3a+\frac{1}{4b}\right)}^{3}$
(iii) ${\left(1+\frac{2}{3}a\right)}^{3}$

#### Question 2:

Expand
(i) (5a – 3b)3
(ii) ${\left(3x-\frac{5}{x}\right)}^{3}$
(iii) ${\left(\frac{4}{5}a-2\right)}^{3}$

#### Question 3:

Factorise
$8{a}^{3}+27{b}^{3}+36{a}^{2}b+54a{b}^{2}$

Hence, factorisation of $8{a}^{3}+27{b}^{3}+36{a}^{2}b+54a{b}^{2}$ is ${\left(2a+3b\right)}^{3}$.

#### Question 4:

Factorise
$64{a}^{3}-27{b}^{3}-144{a}^{2}b+108a{b}^{2}$

Hence, factorisation of $64{a}^{3}-27{b}^{3}-144{a}^{2}b+108a{b}^{2}$ is ${\left(4a-3b\right)}^{3}$.

#### Question 5:

Factorise
$1+\frac{27}{125}{a}^{3}+\frac{9a}{5}+\frac{27{a}^{2}}{24}$

Hence, factorisation of $1+\frac{27}{125}{a}^{3}+\frac{9a}{5}+\frac{27{a}^{2}}{25}$ is ${\left(1+\frac{3}{5}a\right)}^{3}$.

#### Question 6:

Factorise
$125{x}^{3}-27{y}^{3}-225{x}^{2}y+135x{y}^{2}$

Hence, factorisation of $125{x}^{3}-27{y}^{3}-225{x}^{2}y+135x{y}^{2}$ is ${\left(5x-3y\right)}^{3}$.

#### Question 7:

Factorise
${a}^{3}{x}^{3}-3{a}^{2}b{x}^{2}+3a{b}^{2}x-{b}^{3}$

Hence, factorisation of ${a}^{3}{x}^{3}-3{a}^{2}b{x}^{2}+3a{b}^{2}x-{b}^{3}$ is ${\left(ax-b\right)}^{3}$.

#### Question 8:

Factorise
$\frac{64}{125}{a}^{3}-\frac{96}{25}{a}^{2}+\frac{48}{5}a-8$

Hence, factorisation of $\frac{64}{125}{a}^{3}-\frac{96}{25}{a}^{2}+\frac{48}{5}a-8$ is ${\left(\frac{4}{5}a-2\right)}^{3}$.

#### Question 9:

Factorise
a3 – 12a(a – 4) – 64

Hence, factorisation of a3 – 12a(a – 4) – 64 is ${\left(a-4\right)}^{3}$.

Evaluate
(i) (103)3
(ii) (99)3

Factorize:
x3 + 27

#### Question 2:

Factorise
27a3 + 64b3

We know that
${x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}+{y}^{2}-xy\right)$
Given: 27a3 + 64b3
x = 3a, y = 4b

#### Question 3:

Factorize:
$125{a}^{3}+\frac{1}{8}$

#### Question 4:

Factorize:
$216{x}^{3}+\frac{1}{125}$

Factorize:
16x4 + 54x

Factorize:
7a3 + 56b3

Factorize:
x5 + x2

Factorize:
a3 + 0.008

Factorise
1 – 27a3

Factorize:
64a3 − 343

Factorize:
x3 − 512

Factorize:
a3 − 0.064

#### Question 13:

Factorize:
$8{x}^{3}-\frac{1}{27{y}^{3}}$

#### Question 14:

Factorise
$\frac{{x}^{3}}{216}-8{y}^{3}$

We know
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$
We have,
$\frac{{x}^{3}}{216}-8{y}^{3}={\left(\frac{x}{6}\right)}^{3}-{\left(2y\right)}^{3}$
So, $a=\frac{x}{6},b=2y$
$\frac{{x}^{3}}{216}-8{y}^{3}=\left(\frac{x}{6}-2y\right)\left({\left(\frac{x}{6}\right)}^{2}+\frac{x}{6}×2y+{\left(2y\right)}^{2}\right)=\left(\frac{x}{6}-2y\right)\left(\frac{{x}^{2}}{36}+\frac{xy}{3}+4{y}^{2}\right)$

Factorize:
x − 8xy3

Factorise
32x4 – 500x

Factorize:
3a7b − 81a4b4

#### Question 18:

Factorise
x4 y4xy

Using the identity
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$

#### Question 19:

Factorise
8x2 y3x5

$8{x}^{2}{y}^{3}–{x}^{5}={x}^{2}\left(8{y}^{3}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(2y-x\right)\left(4{y}^{2}+{x}^{2}+2xy\right)$

#### Question 20:

Factorise
1029 – 3x3

$1029–3{x}^{3}\phantom{\rule{0ex}{0ex}}=3\left(343-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=3\left({7}^{3}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=3\left(7-x\right)\left(49+{x}^{2}+7x\right)$

Factorize:
x6 − 729

Factorise
x9 – y9

Factorize:
(a + b)3 − (ab)3

#### Question 24:

Factorize:
8a3b3 − 4ax + 2bx

#### Question 25:

Factorize:
a3 + 3a2b + 3ab2 + b3 − 8

#### Question 26:

Factorize:
${a}^{3}-\frac{1}{{a}^{3}}-2a+\frac{2}{a}$

#### Question 27:

Factorize:
2a3 + 16b3 − 5a − 10b

Factorise
a6 + b6

#### Question 29:

Factorise
a12 – b12

a12 – b12
$=\left({a}^{6}+{b}^{6}\right)\left({a}^{6}-{b}^{6}\right)\phantom{\rule{0ex}{0ex}}=\left[{\left({a}^{2}\right)}^{3}+{\left({b}^{2}\right)}^{3}\right]\left[{\left({a}^{3}\right)}^{2}-{\left({b}^{3}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left[\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)\right]\left[\left({a}^{3}-{b}^{3}\right)\left({a}^{3}+{b}^{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\left[\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)\right]\left[\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\right]\phantom{\rule{0ex}{0ex}}=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)\left(a+b\right)\left({a}^{2}+{b}^{2}-ab\right)\left({a}^{2}+{b}^{2}\right)\left({a}^{4}+{b}^{4}-{a}^{2}{b}^{2}\right)$

#### Question 30:

Factorise
x6 – 7x3 – 8

Let ${x}^{3}=y$
So, the equation becomes
${y}^{2}-7y-8={y}^{2}-8y+y-8\phantom{\rule{0ex}{0ex}}=y\left(y-8\right)+\left(y-8\right)\phantom{\rule{0ex}{0ex}}=\left(y-8\right)\left(y+1\right)\phantom{\rule{0ex}{0ex}}=\left({x}^{3}-8\right)\left({x}^{3}+1\right)\phantom{\rule{0ex}{0ex}}=\left(x-2\right)\left({x}^{2}+4+2x\right)\left(x+1\right)\left({x}^{2}+1-x\right)$

#### Question 31:

Factorise
x3 – 3x2 + 3x + 7

x3 – 3x+ 3x + 7
$={x}^{3}–3{x}^{2}+3x+7\phantom{\rule{0ex}{0ex}}={x}^{3}–3{x}^{2}+3x+8-1={x}^{3}–3{x}^{2}+3x-1+8\phantom{\rule{0ex}{0ex}}=\left({x}^{3}–3{x}^{2}+3x-1\right)+8\phantom{\rule{0ex}{0ex}}={\left(x-1\right)}^{3}+{2}^{3}\phantom{\rule{0ex}{0ex}}=\left(x-1+2\right)\left[{\left(x-1\right)}^{2}+4-2\left(x-1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left[{x}^{2}+1-2x+4-2x+2\right]\phantom{\rule{0ex}{0ex}}=\left(x+1\right)\left({x}^{2}-4x+7\right)$

#### Question 32:

Factorise
(x +1)3 + (x – 1)3

(x +1)3 + (x – 1)3
$=\left(x+1+x-1\right)\left[{\left(x+1\right)}^{2}+{\left(x-1\right)}^{2}-\left(x-1\right)\left(x+1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(2x\right)\left[{\left(x+1\right)}^{2}+{\left(x-1\right)}^{2}-\left({x}^{2}-1\right)\right]\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+1+2x+{x}^{2}+1-2x-{x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+3\right)$

#### Question 33:

Factorise
(2a +1)3 + (a – 1)3

(2a +1)3 + (a – 1)3
$=\left(2a+1+a-1\right)\left[{\left(2a+1\right)}^{2}+{\left(a-1\right)}^{2}-\left(2a+1\right)\left(a-1\right)\right]\phantom{\rule{0ex}{0ex}}=\left(3a\right)\left[4{a}^{2}+1+4a+{a}^{2}+1-2a-2{a}^{2}+2a-a+1\right]\phantom{\rule{0ex}{0ex}}=3a\left[3{a}^{2}+3a+3\right]=9a\left({a}^{2}+a+1\right)$

#### Question 34:

Factorise
8(x +y)3 – 27(x y)3

8(x +y)3 – 27(x – y)3
$={\left[2\left(x+y\right)\right]}^{3}-{\left[3\left(x-y\right)\right]}^{3}\phantom{\rule{0ex}{0ex}}=\left(2x+2y-3x+3y\right)\left[4{\left(x+y\right)}^{2}+9{\left(x-y\right)}^{2}+6\left({x}^{2}-{y}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left[4\left({x}^{2}+{y}^{2}+2xy\right)+9\left({x}^{2}+{y}^{2}-2xy\right)+6\left({x}^{2}-{y}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left[4{x}^{2}+4{y}^{2}+8xy+9{x}^{2}+9{y}^{2}-18xy+6{x}^{2}-6{y}^{2}\right]\phantom{\rule{0ex}{0ex}}=\left(-x+5y\right)\left(19{x}^{2}+7{y}^{2}-10xy\right)$

#### Question 35:

Factorise
(x +2)3 + (x – 2)3

(x +2)3 + (x – 2)3
$=\left(x+2+x-2\right)\left[{\left(x+2\right)}^{2}+{\left(x-2\right)}^{2}-\left({x}^{2}-4\right)\right]\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+4+4x+{x}^{2}+4-4x-{x}^{2}+4\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}+12\right)\phantom{\rule{0ex}{0ex}}$

#### Question 36:

Factorise
(x + 2)3 – (x – 2)3

(x + 2)3 – (x – 2)3
$=\left(x+2-x+2\right)\left[{\left(x+2\right)}^{2}+{\left(x-2\right)}^{2}+\left({x}^{2}-4\right)\right]\phantom{\rule{0ex}{0ex}}=4\left[{x}^{2}+4+4x+{x}^{2}+4-4x+{x}^{2}-4\right]\phantom{\rule{0ex}{0ex}}=4\left(3{x}^{2}+4\right)$

#### Question 37:

Prove that $\frac{0.85×0.85×0.85+0.15×0.15×0.15}{0.85×0.85-0.85×0.15+0.15×0.15}=1$.

$\frac{{\left(0.85\right)}^{3}+{\left(0.15\right)}^{3}}{{\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(0.85+0.15\right)\left({\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}\right)}{{\left(0.85\right)}^{2}-0.85×0.15+{\left(0.15\right)}^{2}}\phantom{\rule{0ex}{0ex}}=0.85+0.15=1:\mathrm{RHS}$
Thus, LHS = RHS

#### Question 38:

Prove that $\frac{59×59×59-9×9×9}{59×59+59×9+9×9}=50$.

$\frac{59×59×59-9×9×9}{59×59+59×9+9×9}\phantom{\rule{0ex}{0ex}}=\frac{{\left(59\right)}^{3}-{9}^{3}}{{59}^{2}+59×9+{9}^{2}}\phantom{\rule{0ex}{0ex}}$

Thus, LHS=RHS

#### Question 1:

Find the product:
(x + yz) (x2 + y2 + z2xy + yz + zx)

#### Question 2:

Find the product:
(xyz) (x2 + y2 + z2 + xyyz + xz)

(x – y − z) (x2 + y2 + z2 + xy – yz + xz)

#### Question 3:

Find the product:
(x − 2y + 3) (x2 + 4y2 + 2xy − 3x + 6y + 9)

#### Question 4:

Find the product:
(3x – 5y + 4) (9x2 + 25y2 + 15xy − 20y + 12x + 16)

$\left(3x-5y+4\right)\left(9{x}^{2}+25{y}^{2}+15xy-20y+12x+16\right)\phantom{\rule{0ex}{0ex}}=\left(3x+\left(-5y\right)+4\right)\left(9{x}^{2}+25{y}^{2}+16+15xy-20y+12x\right)$

$\left(3x+\left(-5y\right)+4\right)\left(9{x}^{2}+25{y}^{2}+16+15xy-20y+12x\right)\phantom{\rule{0ex}{0ex}}={\left(3x\right)}^{3}+{\left(-5y\right)}^{3}+{4}^{3}-3×3x\left(-5y\right)\left(4\right)\phantom{\rule{0ex}{0ex}}=27{x}^{3}-125{y}^{3}+64+180xy$

#### Question 5:

Factorize:
125a3 + b3 + 64c3 − 60abc

#### Question 6:

Factorize:
a3 + 8b3 + 64c3 − 24abc

#### Question 7:

Factorize:
1 + b3 + 8c3 − 6bc

#### Question 8:

Factorize:
216 + 27b3 + 8c3 − 108abc

#### Question 9:

Factorize:
27a3b3 + 8c3 + 18abc

#### Question 10:

Factorize:
8a3 + 125b3 − 64c3 + 120abc

#### Question 11:

Factorize:
8 − 27b3 − 343c3 − 126bc

#### Question 12:

Factorize:
125 − 8x3 − 27y3 − 90xy

Factorize:

Factorise:
27x3 y3z3 – 9xyz

#### Question 15:

Factorise:
$2\sqrt{2}{a}^{3}+3\sqrt{3}{b}^{3}+{c}^{3}-3\sqrt{6}abc$

#### Question 16:

Factorise:
$3\sqrt{3}{a}^{3}-{b}^{3}-5\sqrt{5}{c}^{3}-3\sqrt{15}abc$

#### Question 17:

Factorize:
(ab)3 + (bc)3 + (ca)3

${\left(a-b\right)}^{3}+{\left(b-c\right)}^{3}+{\left(c-a\right)}^{3}$

#### Question 18:

Factorise:
${\left(a-3b\right)}^{3}+{\left(3b-c\right)}^{3}+{\left(c-a\right)}^{3}$

${\left(a-3b\right)}^{3}+{\left(3b-c\right)}^{3}+{\left(c-a\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(a-3b+3b-c+c-a\right)\left[{\left(a-3b\right)}^{2}+{\left(3b-c\right)}^{2}+{\left(c-a\right)}^{2}-\left(a-3b\right)\left(3b-c\right)-\left(3b-c\right)\left(c-a\right)-\left(c-a\right)\left(a-3b\right)\right]\phantom{\rule{0ex}{0ex}}+3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)\phantom{\rule{0ex}{0ex}}=0+3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)\phantom{\rule{0ex}{0ex}}=3\left(a-3b\right)\left(3b-c\right)\left(c-a\right)$

#### Question 19:

Factorize:
(3a − 2b)3 + (2b − 5c)3 + (5c − 3a)3

#### Question 20:

Factorize:
(5a − 7b)3 + (9c − 5a)3 + (7b − 9c)3

#### Question 21:

Factorize:
a3(bc)3 + b3(ca)3 + c3(ab)3

#### Question 22:

Evaluate
(i) (–12)3 + 73 + 53
(ii) (28)3 + (–15)3 + (–13)3

(i) (–12)+ 7+ 53

${\left(-12\right)}^{3}+{7}^{3}+{5}^{3}\phantom{\rule{0ex}{0ex}}=\left(-12+7+5\right)\left[{\left(-12\right)}^{2}+{7}^{2}+{5}^{2}-7\left(-12\right)-35+60\right]+3\left(-12\right)×35\phantom{\rule{0ex}{0ex}}=0-1260=-1260$

(ii) (28)3 + (–15)3 + (–13)3

${\left(28\right)}^{3}+{\left(-15\right)}^{3}+{\left(-13\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(28-15-13\right)\left[{\left(28\right)}^{2}+{\left(-15\right)}^{2}+{\left(-13\right)}^{2}-28\left(-15\right)-\left(-15\right)\left(-13\right)-28\left(-13\right)\right]+3×28\left(-15\right)\left(-13\right)\phantom{\rule{0ex}{0ex}}=0+16380=16380$

#### Question 23:

Prove that

${\left(a+b+c\right)}^{3}={\left[\left(a+b\right)+c\right]}^{3}={\left(a+b\right)}^{3}+{c}^{3}+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}={a}^{3}+{b}^{3}+3ab\left(a+b\right)+{c}^{3}+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left[ab+ca+cb+{c}^{2}\right]\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\phantom{\rule{0ex}{0ex}}⇒{\left(a+b+c\right)}^{3}-{a}^{3}+{b}^{3}-{c}^{3}=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Question 24:

If a, b, c are all nonzero and a + b + c = 0, prove that $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}=3$.

$a+b+c=0⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}$
$=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}=3$

#### Question 25:

If a + b + c = 9 and a2 + b2 + c2 = 35, find the value of (a3 + b3 + c3 – 3abc).

a + b + c = 9
$⇒{\left(a+b+c\right)}^{2}={9}^{2}=81\phantom{\rule{0ex}{0ex}}⇒{a}^{2}+{b}^{2}+{c}^{2}+2\left(ab+bc+ca\right)=81\phantom{\rule{0ex}{0ex}}⇒35+2\left(ab+bc+ca\right)=81\phantom{\rule{0ex}{0ex}}⇒\left(ab+bc+ca\right)=23$
We know,
(a+ b3 + c3 – 3abc) = $\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)$
$=\left(9\right)\left(35-23\right)\phantom{\rule{0ex}{0ex}}=108$

#### Question 1:

If (x + 1) is factor of the polynomial (2x2 + kx) then the value of k is
(a) –2
(b) –3
(c) 2
(d) 3

(c) 2

#### Question 2:

The value of (249)2 – (248)2 is
(a) 12
(b) 477
(c) 487
(d) 497

(249)2 – (248)2
We know

Hence, the correct answer is option (d).

#### Question 3:

If $\frac{x}{y}+\frac{y}{x}=-1$, where x ≠ 0 and y ≠ 0, then the value of (x3y3) is
(a) 1
(b) −1
(c) 0
(d) $\frac{1}{2}$

(c) 0

$⇒x$2 + y2 = $-$xy
x2 + y2 + xy = 0

Thus, we have:
$\left({x}^{3}-{y}^{3}\right)=\left(x-y\right)\left({x}^{2}+{y}^{2}+xy\right)$
$=\left(x-y\right)×0\phantom{\rule{0ex}{0ex}}=0$

#### Question 4:

If a + b + c = 0, then a3 + b3 + c3 = ?
(a) 0
(b) abc
(c) 2abc
(d) 3abc

(d) 3abc

$⇒{\left(a+b\right)}^{3}={\left(-c\right)}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(a+b\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+3ab\left(-c\right)=-{c}^{3}\phantom{\rule{0ex}{0ex}}⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

#### Question 5:

If then the value of p is
(a) 0
(b) $-\frac{1}{4}$
(c) $\frac{1}{4}$
(d) $\frac{1}{2}$

Hence, the correct answer is option (c).

#### Question 6:

The coefficient of x in the expansion of (x + 3)3 is
(a) 1
(b) 9
(c) 18
(d) 27

(x + 3)3
$={x}^{3}+{3}^{3}+9x\left(x+3\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+27+9{x}^{2}+27x$
So, the coefficient of x in (x + 3)is 27.
Hence, the correct answer is option (d).

#### Question 7:

Which of the following is a factor of (x + y)3 – (x3 + y3)?
(a) x2 + y2 + 2xy
(b) x2 + y2xy
(c) xy2
(d) 3xy

(x + y)3 – (xy3)
$={x}^{3}+{y}^{3}+3xy\left(x+y\right)-\left({x}^{3}+{y}^{3}\right)\phantom{\rule{0ex}{0ex}}=3xy\left(x+y\right)$
Thus, the factors of (x + y)3 – (xy3) are 3xy and (x + y).
Hence, the correct answer is option (d).

#### Question 8:

One of the factors of $\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}$ is
(a) 5 + x
(b) 5 – x
(c) 5x – 1
(d) 10x

$\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x-1\right)\left(5x+1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left[\left(5x-1\right)+\left(1+5x\right)\right]\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left(10x\right)$
So, the factors of $\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}$ are (5x + 1) and 10x
Hence, the correct answer is option (d).

#### Question 9:

If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3

(b) 5

#### Question 10:

If (x + 2) and (x − 1) are factors of (x3 + 10x2 + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19

(b) m = 7, n = −18

Let:
$p\left(x\right)={x}^{3}+10{x}^{2}+mx+n\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Now,
$x+2=0⇒x=-2$
(x + 2) is a factor of p(x).
So, we have p($-$2)=0

Now,
$x-1=0⇒x=1$
Also,
(x $-$ 1) is a factor of p(x).
We have:
p(1) = 0

By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18

#### Question 11:

104 × 96 = ?
(a) 9864
(b) 9984
(c) 9684
(d) 9884

(b) 9984

$104×96=\left(100+4\right)\left(100-4\right)$
$={100}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}=\left(10000-16\right)\phantom{\rule{0ex}{0ex}}=9984$

305 × 308 = ?
(a) 94940
(b) 93840
(c) 93940
(d) 94840

(c) 93940

207 × 193 = ?
(a) 39851
(b) 39951
(c) 39961
(d) 38951

(b) 39951

#### Question 14:

4a2 + b2 + 4ab + 8a + 4b + 4 = ?
(a) (2a + b + 2)2
(b) (2ab + 2)2
(c) (a + 2b + 2)2
(d) none of these

(a) (2a + b + 2)2

#### Question 15:

(x2 − 4x − 21) = ?
(a) (x − 7)(x − 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x + 3)
(d) none of these

(c) (x − 7)(x + 3)

${x}^{2}-4x-21={x}^{2}-7x+3x-21$
$=x\left(x-7\right)+3\left(x-7\right)\phantom{\rule{0ex}{0ex}}=\left(x-7\right)\left(x+3\right)$

#### Question 16:

(4x2 + 4x − 3) = ?
(a) (2x − 1) (2x − 3)
(b) (2x + 1) (2x − 3)
(c) (2x + 3) (2x − 1)
(d) none of these

(c) (2x + 3) (2x − 1)

$4{x}^{2}+4x-3=4{x}^{2}+6x-2x-3$
$=2x\left(2x+3\right)-1\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=\left(2x+3\right)\left(2x-1\right)$

#### Question 17:

6x2 + 17x + 5 = ?
(a) (2x + 1)(3x + 5)
(b) (2x + 5)(3x + 1)
(c) (6x + 5)(x + 1)
(d) none of these

(b) (2x + 5)(3x + 1)

$6{x}^{2}+17x+5=6{x}^{2}+15x+2x+5$
$=3x\left(2x+5\right)+1\left(2x+5\right)\phantom{\rule{0ex}{0ex}}=\left(2x+5\right)\left(3x+1\right)$

#### Question 18:

(x + 1) is a factor of the polynomial
(a) x3 − 2x2 + x + 2
(b) x3 + 2x2 + x − 2
(c) x3 − 2x2 − x − 2
(d) x3 − 2x2 − x + 2

(c) x3 − 2x2 − x − 2

Let:
$f\left(x\right)={x}^{3}-2{x}^{2}+x+2$
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}-2{x}^{2}+x+2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}+x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is not a factor of $f\left(x\right)={x}^{3}+2{x}^{2}+x-2$.

Now,
Let:
$f\left(x\right)={x}^{3}+2{x}^{2}-x-2$

By the factor theorem, (x + 1) will be a factor of f (x) if f ($-$1) = 0.
We have:

Hence, (x + 1) is a factor of $f\left(x\right)={x}^{3}+2{x}^{2}-x-2$.

#### Question 19:

3x3 + 2x2 + 3x + 2 = ?
(a) (3x − 2)(x2 − 1)
(b) (3x − 2)(x2 + 1)
(c) (3x + 2)(x2 − 1)
(d) (3x + 2)(x2 + 1)

(d) (3x + 2)(x2 + 1)

$3{x}^{3}+2{x}^{2}+3x+2={x}^{2}\left(3x+2\right)+1\left(3x+2\right)$
$=\left(3x+2\right)\left({x}^{2}+1\right)$

#### Question 20:

If a + b + c = 0, then $\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)$=?

(d) 3

$a+b+c=0⇒{a}^{3}+{b}^{3}+{c}^{3}=3abc$

Thus, we have:
$\left(\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}\right)=\frac{{a}^{3}+{b}^{3}+{c}^{3}}{abc}$
$=\frac{3abc}{abc}\phantom{\rule{0ex}{0ex}}=3$

#### Question 21:

If x + y + z = 9 and xy + yz + zx = 23, then the value of (x3 + y3 + z3 − 3xyz) = ?
(a) 108
(b) 207
(c) 669
(d) 729

(a) 108

${x}^{3}+{y}^{3}+{z}^{3}-3xyz=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx\right)\phantom{\rule{0ex}{0ex}}$
$=\left(x+y+z\right)\left[{\left(x+y+z\right)}^{2}-3\left(xy+yz+zx\right)\right]\phantom{\rule{0ex}{0ex}}=9×\left(81-3×23\right)\phantom{\rule{0ex}{0ex}}=9×12\phantom{\rule{0ex}{0ex}}=108$

#### Question 22:

If $\frac{a}{b}+\frac{b}{a}=-1$ then (a3b3) = ?
(a) −3
(b) −2
(c) −1
(d) 0

$⇒a$2 + b2 = $-$ab
$⇒a$2 + b2 + ab = 0
$\left({a}^{3}-{b}^{3}\right)=\left(a-b\right)\left({a}^{2}+{b}^{2}+ab\right)$
$=\left(a-b\right)×0\phantom{\rule{0ex}{0ex}}=0$