Rs Aggarwal 2020 2021 Solutions for Class 9 Maths 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 Maths Factorisation Of Polynomials Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 9 Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

We have:

#### Page No 99:

We have:

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

#### Page No 99:

We have:

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 99:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 100:

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

#### Page No 105:

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

#### Page No 105:

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.

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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,

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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, .

#### Page No 114:

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, .

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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

#### Page No 114:

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:

#### Page No 114:

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:

#### Page No 115:

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

#### Page No 115:

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

#### Page No 115:

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

#### Page No 115:

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

#### Page No 115:

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

#### Page No 115:

Hence, {(999)2 – 1} = 998000.

#### Page No 119:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 123:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

#### Page No 129:

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

Thus, LHS=RHS

#### Page No 136:

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

#### Page No 136:

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

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

#### Page No 136:

#### Page No 137:

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

#### Page No 137:

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

#### Page No 137:

(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

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

#### Page No 137:

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

#### Page No 137:

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

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

#### Page No 137:

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) = $\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)$
$=\left(9\right)\left(35-23\right)\phantom{\rule{0ex}{0ex}}=108$

(c) 2

#### Page No 138:

(249)2 – (248)2
We know

Hence, the correct answer is option (d).

#### Page No 138:

(c) 0

$⇒x$2 + y2 = $-$xy
x2 + y2 + xy = 0

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

#### Page No 138:

(d) 3abc

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

#### Page No 139:

Hence, the correct answer is option (c).

#### Page No 139:

(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).

#### Page No 139:

(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).

#### Page No 139:

$\left(25{x}^{2}-1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x-1\right)\left(5x+1\right)+{\left(1+5x\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left[\left(5x-1\right)+\left(1+5x\right)\right]\phantom{\rule{0ex}{0ex}}=\left(5x+1\right)\left(10x\right)$
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).

(b) 5

#### Page No 139:

(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

#### Page No 139:

(b) 9984

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

(c) 93940

(b) 39951

#### Page No 139:

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

#### Page No 139:

(c) (x − 7)(x + 3)

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

#### Page No 139:

(c) (2x + 3) (2x − 1)

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

#### Page No 139:

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

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

#### Page No 139:

(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 $f\left(x\right)={x}^{3}+2{x}^{2}-x-2$.

#### Page No 140:

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

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

#### Page No 140:

(d) 3

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

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

#### Page No 140:

(a) 108

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

#### Page No 140:

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