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#### Page No 99:

(i) $12x+15=3\left(4x+5\right)$
(ii) $14m-21=7\left(2m-3\right)$
(iii) $9n-12{n}^{2}=3n\left(3-4n\right)$

#### Page No 99:

(i) H.C.F. of $16{a}^{2}$ and $24ab$ is $8a$.

∴ $16{a}^{2}-24ab=8a\left(2a-3b\right)$

(ii) H.C.F. of $15a{b}^{2}$ and $20{a}^{2}b$ is $5ab$.

∴ $15a{b}^{2}-20{a}^{2}b=5ab\left(3b-4a\right)$

(iii) ​H.C.F. of $12{x}^{2}{y}^{3}$ and $21{x}^{3}{y}^{2}$ is $3{x}^{2}{y}^{2}$.

∴ $12{x}^{2}{y}^{3}-21{x}^{3}{y}^{2}=3{x}^{2}{y}^{2}\left(4y-7x\right)$

#### Page No 99:

(i) H.C.F. of $24{x}^{3}$ and $36{x}^{2}y$ is $12{x}^{2}$.

∴  $24{x}^{3}-36{x}^{2}y=12{x}^{2}\left(2x-3y\right)$

(ii)  H.C.F. of $10{x}^{3}$ and $15{x}^{2}$ is $5{x}^{3}$.

∴ $10{x}^{3}-15{x}^{2}=5{x}^{2}\left(2x-3\right)$

(iii) H.C.F. of $36{x}^{3}y$ and $60{x}^{2}{y}^{3}z$ is $12{x}^{2}y$.

∴ $36{x}^{3}y-60{x}^{2}{y}^{3}z=12{x}^{2}y\left(3x-5{y}^{2}z\right)$

#### Page No 99:

(i) H.C.F. of $9{x}^{3}$$6{x}^{2}$ and $12x$ is $3x$.

∴ $9{x}^{3}-6{x}^{2}+12x=3x\left(3{x}^{2}-2x+4\right)$

(ii) H.C.F. of $8{x}^{3}$$72xy$ and $12x$ is $4x$.

∴  $8{x}^{3}-72xy+12x=4x\left(2{x}^{2}-18y+3\right)$

(iii) H.C.F. of $18{a}^{3}{b}^{3}$$27{a}^{2}{b}^{3}$ and $36{a}^{3}{b}^{2}$ is $9{a}^{2}{b}^{2}$.

∴ $18{a}^{3}{b}^{3}-27{a}^{2}{b}^{3}+36{a}^{3}{b}^{2}=9{a}^{2}{b}^{2}\left(2ab-3b+4a\right)$

#### Page No 99:

(i) H.C.F. of $14{x}^{3}$$21{x}^{4}y$ and $28{x}^{2}{y}^{2}$ is $7{x}^{2}$.

∴ $14{x}^{3}+21{x}^{4}y-28{x}^{2}{y}^{2}=7{x}^{2}\left(2x+3{x}^{2}y-4{y}^{2}\right)$

(ii) H.C.F. of $-5$$-10t$ and $20{t}^{2}$ is 5.

∴ $-5-10t+20{t}^{2}=5\left(-1-2t+4{t}^{2}\right)$

#### Page No 99:

(i) $x\left(x+3\right)+5\left(x+3\right)=\left(x+3\right)\left(x+5\right)$

(ii) $5x\left(x-4\right)-7\left(x-4\right)=\left(x-4\right)\left(5x-7\right)$

(iii) $2m\left(1-n\right)+3\left(1-n\right)=\left(1-n\right)\left(2m+3\right)$

#### Page No 99:

We have:
$6a\left(a-2b\right)+5b\left(a-2b\right)=\left(a-2b\right)\left(6a+5b\right)$

#### Page No 99:

We have:
${x}^{3}\left(2a-b\right)+{x}^{2}\left(2a-b\right)=\left(2a-b\right)\left({x}^{3}+{x}^{2}\right)={x}^{2}\left(x+1\right)\left(2a-b\right)$

#### Page No 99:

We have:
$9a\left(3a-5b\right)-12{a}^{2}\left(3a-5b\right)=\left(3a-5b\right)\left(9a-12{a}^{2}\right)=3a\left(3a-5b\right)\left(3-4a\right)$

#### Page No 99:

We have:
${\left(x+5\right)}^{2}-4\left(x+5\right)=\left(x+5\right)\left\{\left(x+5\right)-4\right\}$
$=\left(x+5\right)\left(x+5-4\right)\phantom{\rule{0ex}{0ex}}=\left(x+5\right)\left(x+1\right)$

${\left(x+5\right)}^{2}-4\left(x+5\right)=\left(x+5\right)\left(x+1\right)$

#### Page No 99:

$3{\left(a-2b\right)}^{2}-5\left(a-2b\right)=\left(a-2b\right)\left\{3\left(a-2b\right)-5\right\}$
$=\left(a-2b\right)\left(3a-6b-5\right)$

∴ $3{\left(a-2b\right)}^{2}-5\left(a-2b\right)=\left(a-2b\right)\left(3a-6b-5\right)$

#### Page No 99:

We have:

$2a+6b-3{\left(a+3b\right)}^{2}=\left(a+3b\right)\left(2-3a-9b\right)$

#### Page No 99:

We have:

∴ $16{\left(2p-3q\right)}^{2}-4\left(2p-3q\right)=\left(2p-3q\right)\left(32p-48q-4\right)$

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∴

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#### Page No 99:

By grouping the terms:

#### Page No 99:

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#### Page No 99:

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#### Page No 99:

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#### Page No 99:

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#### Page No 99:

We have:

∴ $ab\left({x}^{2}+{y}^{2}\right)-xy\left({a}^{2}+{b}^{2}\right)=\left(bx-ay\right)\left(ax-by\right)$

#### Page No 99:

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

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

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#### Page No 100:

We have:

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

∴

#### Page No 100:

We have:
$25-{a}^{2}-{b}^{2}-2ab=25-\left({a}^{2}+{b}^{2}+2ab\right)$
$=25-{\left(a+b\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(5\right)}^{2}-{\left(a+b\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left\{5+\left(a+b\right)\right\}\left\{5-\left(a+b\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(5+a+b\right)\left(5-a-b\right)$

∴ $25-{a}^{2}-{b}^{2}-2ab=\left(5+a+b\right)\left(5-a-b\right)$

#### Page No 100:

We have:
$25{a}^{2}-4{b}^{2}+28bc-49{c}^{2}=25{a}^{2}-\left(4{b}^{2}-28bc+49{c}^{2}\right)$
$={\left(5a\right)}^{2}-{\left(2b-7c\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left\{5a+\left(2b-7c\right)\right\}\left\{5a-\left(2b-7c\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(5a+2b-7c\right)\left(5a-2b+7c\right)$

∴ $25{a}^{2}-4{b}^{2}+28bc-49{c}^{2}=\left(5a+2b-7c\right)\left(5a-2b+7c\right)$

#### Page No 100:

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

∴ $9{a}^{2}-{b}^{2}+4b-4=\left(3a+b-2\right)\left(3a-b+2\right)$

#### Page No 100:

We have:
$100-{\left(x-5\right)}^{2}={\left(10\right)}^{2}-{\left(x-5\right)}^{2}$
$=\left\{10+\left(x-5\right)\right\}\left\{10-\left(x-5\right)\right\}\phantom{\rule{0ex}{0ex}}=\left(10+x-5\right)\left(10-x+5\right)\phantom{\rule{0ex}{0ex}}=\left(5+x\right)\left(15-x\right)$

∴ $100-{\left(x-5\right)}^{2}=\left(5+x\right)\left(15-x\right)$

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∴

#### Page No 100:

We have:

∴ $\left\{{\left(7.8\right)}^{2}-{\left(2.2\right)}^{2}\right\}=56$

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#### Page No 101:

We have:

$={\left(3m+4\right)}^{2}$

∴

#### Page No 101:

We have:
${z}^{2}+z+\frac{1}{4}={z}^{2}+2×z×\frac{1}{2}×{\left(\frac{1}{2}\right)}^{2}$
$={\left(z+\frac{1}{2}\right)}^{2}$

∴ ${z}^{2}+z+\frac{1}{4}={\left(z+\frac{1}{2}\right)}^{2}$

#### Page No 101:

We have:
$49{a}^{2}+84ab+36{b}^{2}={\left(7a\right)}^{2}+2×7a×6b+{\left(6b\right)}^{2}$
$={\left(7a+6b\right)}^{2}$

∴ $49{a}^{2}+84ab+36{b}^{2}={\left(7a+6b\right)}^{2}$

#### Page No 101:

We have:
${p}^{2}-10p+25={p}^{2}-2×p×5+{\left(5\right)}^{2}$
$={\left(p-5\right)}^{2}$

∴ ${p}^{2}-10p+25={\left(p-5\right)}^{2}$

#### Page No 101:

We have:
$121{a}^{2}-88ab+16{b}^{2}={\left(11a\right)}^{2}-2×11a×4b+{\left(4b\right)}^{2}$
$={\left(11a-4b\right)}^{2}$

∴ $121{a}^{2}-88ab+16{b}^{2}={\left(11a-4b\right)}^{2}$

#### Page No 101:

We have:
$1-6x+9{x}^{2}=9{x}^{2}-6x+1$
$={\left(3x\right)}^{2}-2×3x×1+{\left(1\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(3x-1\right)}^{2}$

∴ $1-6x+9{x}^{2}={\left(3x-1\right)}^{2}$

#### Page No 101:

We have:
$9{y}^{2}-12y+4={\left(3y\right)}^{2}-2×3y×2+{\left(2\right)}^{2}$
$={\left(3y-2\right)}^{2}$

∴ $9{y}^{2}-12y+4={\left(3y-2\right)}^{2}$

#### Page No 101:

We have:
$16{x}^{2}-24x+9={\left(4x\right)}^{2}-2×4x×3+{\left(3\right)}^{2}$
$={\left(4x-3\right)}^{2}$

∴ $16{x}^{2}-24x+9={\left(4x-3\right)}^{2}$

#### Page No 101:

We have:
${m}^{2}-4mn+4{n}^{2}={m}^{2}-2×m×2n+{\left(2n\right)}^{2}$
$={\left(m-2n\right)}^{2}$

∴ ${m}^{2}-4mn+4{n}^{2}={\left(m-2n\right)}^{2}$

#### Page No 101:

We have:
${a}^{2}{b}^{2}-6abc+9{c}^{2}={\left(ab\right)}^{2}-2×ab×3c+{\left(3c\right)}^{2}$
$={\left(ab-3c\right)}^{2}$

#### Page No 101:

We have:
${m}^{4}+2{m}^{2}{n}^{2}+{n}^{4}={\left({m}^{2}\right)}^{2}+2×{m}^{2}×{n}^{2}+{\left({n}^{2}\right)}^{2}$
$={\left({m}^{2}+{n}^{2}\right)}^{2}$

∴ ${m}^{4}+2{m}^{2}{n}^{2}+{n}^{4}={\left({m}^{2}+{n}^{2}\right)}^{2}$

#### Page No 101:

We have:
${\left(l+m\right)}^{2}-4lm=\left({l}^{2}+{m}^{2}+2lm\right)-4lm$
$={l}^{2}+{m}^{2}+2lm-4lm\phantom{\rule{0ex}{0ex}}={l}^{2}+{m}^{2}-2lm\phantom{\rule{0ex}{0ex}}={\left(l\right)}^{2}+{\left(m\right)}^{2}-2×l×m\phantom{\rule{0ex}{0ex}}={\left(l-m\right)}^{2}$

∴ ${\left(l+m\right)}^{2}-4lm={\left(l-m\right)}^{2}$

#### Page No 103:

(d) 7(a − 3b)(a + 3b)

#### Page No 103:

(d) 2x(1 − 4x)(1 + 4x)

$\left(2x-32{x}^{3}\right)\phantom{\rule{0ex}{0ex}}=2x\left(1-16{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=2x\left(1-4x\right)\left(1+4x\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 103:

(c) x(x − 12)(x + 12)

#### Page No 103:

(d) 2(1 − 5x)(1 + 5x)

#### Page No 103:

(a) (a + b)(a + c)

#### Page No 103:

(d) (pq − 1) (q + 1)

$p{q}^{2}+q\left(p-1\right)-1\phantom{\rule{0ex}{0ex}}=p{q}^{2}+qp-q-1\phantom{\rule{0ex}{0ex}}=pq\left(q+1\right)-1\left(q+1\right)\phantom{\rule{0ex}{0ex}}=\left(pq-1\right)\left(q+1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 103:

(b) (am)(b + n)

$ab-mn+an-bm\phantom{\rule{0ex}{0ex}}=ab+an-mn-bm\phantom{\rule{0ex}{0ex}}=a\left(b+n\right)-m\left(n+b\right)\phantom{\rule{0ex}{0ex}}=\left(a-m\right)\left(b+n\right)$

#### Page No 103:

(a) (a − 1)(b − 1)

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

(c) (x + y)(xz)

#### Page No 104:

(c) 3(2m − 3)(2m + 3)

#### Page No 104:

(d) x(x + 1)(x − 1)

#### Page No 104:

(c) (1 + a + b)(1 − ab)

#### Page No 104:

(c) (x + 2)(x + 4)

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

#### Page No 104:

(b) (x + 7)(x − 3)

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

#### Page No 104:

(a) (y − 1)(y + 3)

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

#### Page No 104:

(c) (5 + x)(8 − x)

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

#### Page No 104:

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

#### Page No 104:

(c) (3a − 2)(2a − 3)

$6{a}^{2}-13a+6\phantom{\rule{0ex}{0ex}}=6{a}^{2}-9a-4a+6\phantom{\rule{0ex}{0ex}}=3a\left(2a-3\right)-2\left(2a-3\right)\phantom{\rule{0ex}{0ex}}=\left(3a-2\right)\left(2a-3\right)\phantom{\rule{0ex}{0ex}}$

#### Page No 104:

(a) (2z − 1)(2z − 3)

$4{z}^{2}-8z+3\phantom{\rule{0ex}{0ex}}=4{z}^{2}-6z-2z+3\phantom{\rule{0ex}{0ex}}=2z\left(2z-3\right)-1\left(2z-3\right)\phantom{\rule{0ex}{0ex}}=\left(2z-1\right)\left(2z-3\right)$

#### Page No 104:

$3+23y-8{y}^{2}\phantom{\rule{0ex}{0ex}}=3+24y-y-8{y}^{2}\phantom{\rule{0ex}{0ex}}=3\left(1+8y\right)-y\left(1+8y\right)\phantom{\rule{0ex}{0ex}}=\left(3-y\right)\left(1+8y\right)\phantom{\rule{0ex}{0ex}}$