Rs Aggarwal 2019 Solutions for Class 8 Math Chapter 7 Factorisation are provided here with simple step-by-step explanations. These solutions for Factorisation are extremely popular among Class 8 students for Math Factorisation Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 8 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Factorise:
(i) 12x + 15
(ii) 14m − 21
(iii) 9n − 12n2

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

#### Question 2:

Factorise:
(i) 16a2 − 24ab
(ii) 15ab2 − 20a2b
(iii) 12x2y3 − 21x3y2

(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 $3{x}^{2}{y}^{2}$.

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

#### Question 3:

Factorise:
(i) 24x3 − 36x2y
(ii) 10x3 − 15x2
(iii) 36x3y − 60x2y3z

(i) H.C.F. of $24{x}^{3}$ and $36{x}^{2}y$ is $6{x}^{2}$.

∴  $24{x}^{3}-36{x}^{2}y=6{x}^{2}\left(4x-6y\right)$

(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 $12{x}^{2}y$.

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

#### Question 4:

Factorise:
(i) 9x3 − 6x2 + 12x
(ii) 8x2 − 72xy + 12x
(iii) 18a3b3 − 27a2b3 + 36a3b2

(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 $9{a}^{2}{b}^{2}$.

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

#### Question 5:

Factorise:
(i) 14x3 + 21x4y − 28x2y2
(ii) −5 − 10t + 20t2

(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.

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

#### Question 6:

Factorise:
(i) x(x + 3) + 5(x + 3)
(ii) 5x(x − 4) − 7(x − 4)
(iii) 2m(1 − n) + 3(1 − n)

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

#### Question 7:

Factorise:
6a(a − 2b) + 5b(a − 2b)

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

#### Question 8:

Factorise:
x3(2ab) + x2(2ab)

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

#### Question 9:

Factorise:
9a(3a − 5b) − 12a2(3a − 5b)

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

#### Question 10:

Factorise:
(x + 5)2 − 4(x + 5)

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

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

#### Question 11:

Factorise:
3(a − 2b)2 − 5(a − 2b)

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

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

#### Question 12:

Factorise:
2a + 6b − 3(a + 3b)2

We have:

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

#### Question 13:

Factorise:
16(2p − 3q)2 − 4(2p − 3q)

We have:

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

#### Question 14:

Factorise:
x(a − 3) + y(3 − a)

We have:

∴

#### Question 15:

Factorise:
12(2x − 3y)2 − 16(3y − 2x)

We have:

#### Question 16:

Factorise:
(x + y)(2x + 5) − (x + y)(x + 3)

We have:

#### Question 17:

Factorise:
ar + br + at + bt

By grouping the terms:

#### Question 18:

Factorise:
x2axbx + ab

By suitably arranging the terms:

∴

#### Question 19:

Factorise:
ab2bc2ab + c2

By suitably arranging the terms:

∴

#### Question 20:

Factorise:
x2xz + xyyz

By suitably arranging the terms:

∴

#### Question 21:

Factorise:
6abb2 + 12ac − 2bc

By suitably arranging the terms:

∴

#### Question 22:

Factorise:
(x − 2y)2 + 4x − 8y

We have:

∴

Factorise:
y2xy(1 − x) − x3

We have:

∴

#### Question 24:

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

We have:

∴

#### Question 25:

Factorise:
ab2 + (a − 1)b − 1

We have:

∴

Factorise:
x3 − 3x2 + x − 3

We have:

∴

#### Question 27:

Factorise:
ab(x2 + y2) − xy(a2 + b2)

We have:

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

#### Question 28:

Factorise:
x2x(a + 2b) + 2ab

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

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

Factorise:
x2 − 36

We have:

∴

Factorise:
4a2 − 9

We have:

∴

Factorise:
81 − 49x2

We have:

∴

Factorise:
4x2 − 9y2

We have:

∴

Factorise:
16a2 − 225b2

We have:

∴

Factorise:
9a2b2 − 25

We have:

∴

Factorise:
16a2 − 144

We have:

∴

Factorise:
63a2 − 112b2

We have:

∴

Factorise:
20a2 − 45b2

We have:

∴

Factorise:
12x2 − 27

We have:

∴

Factorise:
x3 − 64x

We have:

∴

Factorise:
16x5 − 144x3

We have:

∴

Factorise:
3x5 − 48x3

We have:

∴

Factorise:
16p3 − 4p

We have:

∴

Factorise:
63a2b2 − 7

We have:

∴

Factorise:
1 − (b − c)2

We have:

∴

#### Question 17:

Factorise:
(2a + 3b)2 − 16c2

We have:

∴

Factorise:
(l + m)2 − (lm)2

We have:

∴

Factorise:
(2x + 5y)2 − 1

We have:

∴

Factorise:
36c2 − (5a + b)2

We have:

∴

#### Question 21:

Factorise:
(3x − 4y)2 − 25z2

We have:

∴

#### Question 22:

Factorise:
x2y2 − 2y − 1

We have:

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

∴

#### Question 23:

Factorise:
25 − a2b2 − 2ab

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

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

#### Question 24:

Factorise:
25a2 − 4b2 + 28bc − 49c2

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

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

#### Question 25:

Factorise:
9a2b2 + 4b − 4

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

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

#### Question 26:

Factorise:
100 − (x − 5)2

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

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

#### Question 27:

Factorise:
Evaluate {(405)2 − (395)2}.

We have:

∴

#### Question 28:

Factorise:
Evaluate {(7.8)2 − (2.2)2}.

We have:

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

Factorise:
x2 + 8x + 16

We have:

∴

Factorise:
x2 + 14x + 49

We have:

∴

Factorise:
1 + 2x + x2

We have:

∴

Factorise:
9 + 6z + z2

We have;

∴

Factorise:
x2 + 6ax + 9a2

We have:

∴

Factorise:
4y2 + 20y + 25

We have:

∴

Factorise:
36a2 + 36a + 9

We have:

∴

#### Question 8:

Factorise:
9m2 + 24m + 16

We have:

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

∴

#### Question 9:

Factorise:
z2 + z + $\frac{1}{4}$

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

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

#### Question 10:

Factorise:
49a2 + 84ab + 36b2

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

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

#### Question 11:

Factorise:
p2 − 10p + 25

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

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

#### Question 12:

Factorise:
121a2 − 88ab + 16b2

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

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

#### Question 13:

Factorise:
1 − 6x + 9x2

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

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

#### Question 14:

Factorise:
9y2 − 12y + 4

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

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

#### Question 15:

Factorise:
16x2 − 24x + 9

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

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

#### Question 16:

Factorise:
m2 − 4mn + 4n2

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

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

#### Question 17:

Factorise:
a2b2 − 6abc + 9c2

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

#### Question 18:

Factorise:
m4 + 2m2n2 + n4

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

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

#### Question 19:

Factorise:
(l + m)2 − 4lm

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

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

Factorise:
x2 + 5x + 6

Factorise:
y2 + 10y + 24

Factorise:
z2 + 12z + 27

Factorise:
p2 + 6p + 8

Factorise:
x2 + 15x + 56

Factorise:
y2 + 19y + 60

Factorise:
x2 + 13x + 40

Factorise:
q2 − 10q + 21

Factorise:
p2 + 6p − 16

Factorise:
x2 − 10x + 24

Factorise:
x2 − 23x + 42

Factorise:
x2 − 17x + 16

Factorise:
y2 − 21y + 90

Factorise:
x2 − 22x + 117

Factorise:
x2 − 9x + 20

Factorise:
x2 + x − 132

Factorise:
x2 + 5x − 104

Factorise:
y2 + 7y − 144

Factorise:
z2 + 19z − 150

Factorise:
y2 + y − 72

Factorise:
a2 + 6a − 91

Factorise:
p2 − 4p − 77

Factorise:
x2 − 7x − 30

Factorise:
x2 − 11x − 42

Factorise:
x2 − 5x − 24

Factorise:
y2 − 6y − 135

Factorise:
z2 − 12z − 45

Factorise:
x2 − 4x − 12

Factorise:
3x2 + 10x + 8

Factorise:
3y2 + 14y + 8

Factorise:
3z2 − 10z + 8

Factorise:
2x2 + x − 45

Factorise:
6p2 + 11p − 10

Factorise:
2x2 − 17x − 30

Factorise:
7y2 − 19y − 6

Factorise:
28 − 31x − 5x2

Factorise:
3 + 23z − 8z2

Factorise:
6x2 − 5x − 6

Factorise:
3m2 + 24m + 36

Factorise:
4n2 − 8n + 3

Factorise:
6x2 − 17x − 3

Factorise:
7x2 − 19x − 6

#### Question 1:

Tick (✓) the correct answer:
(7a2 − 63b2) = ?
(a) (7a − 9b)(9a + 7b)
(b) (7a − 9b)(7a + 9b)
(c) 9(a − 3b)(a + 3b)
(d) 7(a − 3b)(a + 3b)

(d) 7(a − 3b)(a + 3b)

#### Question 2:

Tick (✓) the correct answer:
(2x − 32x3) = ?
(a) 2(x − 4)(x + 4)
(b) 2x(1 − 2x)2
(c) 2x(1 + 2x)2
(d) 2x(1 − 4x)(1 + 4x)

(d) 2x(1 − 4x)(1 + 4x)

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

#### Question 3:

Tick (✓) the correct answer:
x3 − 144x = ?
(a) x(x − 12)2
(b) x(x + 12)2
(c) x(x − 12)(x + 12)
(d) none of these

(c) x(x − 12)(x + 12)

#### Question 4:

Tick (✓) the correct answer:
(2 − 50x2) = ?
(a) 2(1 − 5x)2
(b) 2(1 + 5x)2
(c) (2 − 5x)(2 + 5x)
(d) 2(1 − 5x)(1 + 5x)

(d) 2(1 − 5x)(1 + 5x)

#### Question 5:

Tick (✓) the correct answer:
a2 + bc + ab + ac = ?
(a) (a + b)(a + c)
(b) (a + b)(b + c)
(c) (b + c)(c + a)
(d) a(a + b + c)

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

#### Question 6:

Tick (✓) the correct answer:
pq2 + q(p − 1) − 1 = ?
(a) (pq + 1)(q − 1)
(b) p(q + 1)(q − 1)
(c) q(p − 1)(q + 1)
(d) (pq − 1) (q + 1)

(d) (pq − 1) (q + 1)

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

#### Question 7:

Tick (✓) the correct answer:
abmn + anbm = ?
(a) (ab)(mn)
(b) (am)(b + n)
(c) (an)(m + b)
(d) (ma)(nb)

(b) (am)(b + n)

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

#### Question 8:

Tick (✓) the correct answer:
abab + 1 = ?
(a) (a − 1)(b − 1)
(b) (1 − a)(1 − b)
(c) (a − 1)(1 − b)
(d) (1 − a)(b − 1)

(a) (a − 1)(b − 1)

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

#### Question 9:

Tick (✓) the correct answer:
x2xz + xyyz = ?
(a) (xy)(x + z)
(b) (xy)(x z)
(c) (x + y)(xz)
(d) (xy)(z − x)

(c) (x + y)(xz)

#### Question 10:

Tick (✓) the correct answer:
(12m2 − 27) = ?
(a) (2m − 3)(3m − 9)
(b) 3(2m − 9)(3m − 1)
(c) 3(2m − 3)(2m + 3)
(d) none of these

(c) 3(2m − 3)(2m + 3)

#### Question 11:

Tick (✓) the correct answer:
x3x = ?
(a) x(x2x)
(b) x(xx2)
(c) x(1 + x)(1 − x)
(d) x(x + 1)(x − 1)

(d) x(x + 1)(x − 1)

#### Question 12:

Tick (✓) the correct answer:
1 − 2ab − (a2 + b2) = ?
(a) (1 + ab)(1 + a + b)
(b) (1 + a + b)(1 − a + b)
(c) (1 + a + b)(1 − ab)
(d) (1 + a b)(1 − a + b)

(c) (1 + a + b)(1 − ab)

#### Question 13:

Tick (✓) the correct answer:
x2 + 6x + 8 = ?
(a) (x + 3)(x + 5)
(b) (x + 3)(x + 4)
(c) (x + 2)(x + 4)
(d) (x + 1)(x + 8)

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

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

#### Question 14:

Tick (✓) the correct answer:
x2 + 4x − 21 = ?
(a) (x − 7)(x + 3)
(b) (x + 7)(x − 3)
(c) (x − 7)(x − 3)
(d) (x + 7)(x + 3)

(b) (x + 7)(x − 3)

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

#### Question 15:

Tick (✓) the correct answer:
y2 + 2y − 3 = ?
(a) (y − 1)(y + 3)
(b) (y + 1)(y − 3)
(c) (y − 1)(y − 3)
(d) (y + 2)(y − 3)

(a) (y − 1)(y + 3)

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

#### Question 16:

Tick (✓) the correct answer:
40 + 3xx2 = ?
(a) (x + 5)(x − 8)
(b) (5 − x)(8 + x)
(c) (5 + x)(8 − x)
(d) (5 − x)(8 − x)

(c) (5 + x)(8 − x)

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

#### Question 17:

Tick (✓) the correct answer:
2x2 + 5x + 3 = ?
(a) (x + 3)(2x + 1)
(b) (x + 1)(2x + 3)
(c) (2x + 5)(x − 3)
(d) none of these

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

#### Question 18:

Tick (✓) the correct answer:
6a2 − 13a + 6 = ?
(a) (2a + 3)(3a − 2)
(b) (2a − 3)(3a + 2)
(c) (3a − 2)(2a − 3)
(d) (3a + 1)(2a − 3)

(c) (3a − 2)(2a − 3)

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

#### Question 19:

Tick (✓) the correct answer:
4z2 − 8z + 3 = ?
(a) (2z − 1)(2z − 3)
(b) (2z + 1)(3 − 2z)
(c) (2z + 3)(3z − 1)
(c) (z − 1)(4z − 3)

(a) (2z − 1)(2z − 3)

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

#### Question 20:

Tick (✓) the correct answer:
3 + 23y − 8y2 = ?
(a) (1 − 8y)(3 + y)
(b) (1 + 8y)(3 − y)
(c) (1 − 8y)(y − 3)
(d) (8y − 1)(y + 3)

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