NCERT Solutions for Class 8 Maths Chapter 12 Factorisation

NCERT
SOLUTIONS
CLASS - 8TH

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Book : Mathematics Ncert Solutions | Chapter-14 Maths

Class : 8th
Subject : Maths
Chapter Name : Factorisation
Chapter : 14

Exercise - 14.1

Q1 Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p q 2 2

(iv) 2x, 3x , 4 2

(v) 6abc, 24ab , 12a b 2 2

(vi) 16x , −4x , 32x 3 2

(vii) 10pq, 20qr, 30rp
(viii) 3x y , 10x y , 6x y z
2 3 3 2 2 2

Answer. (i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3

The common factors are 2,2,3
And 2 × 2 × 3 = 12

(ii) 2y = 2 × y
22xy = 2 × 11 × x × y

The common Factors are 2,y
And 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q
2 2
28p q = 2 × 2 × 7 × p × p × q × q

The common factors are 2, 7, p, q
And 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x
2
3x = 3 × x × x

4 = 2 × 2

The common factor is 1

(v) 6abc = 2 × 3 × a × b × c
2
24ab = 2 × 2 × 2 × 3 × a × b × b
2
12a b = 2 × 2 × 3 × a × a × b

The common factors are 2, 3, a, b
And 2 × 3 × a × b = 6ab

(vi) 3x y 2 3
= 3 × x × x × y × y × y
2
−4x = −1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

Page 1 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
The common factors are 2, 2, x
And 2 × 5 = 10

(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

The common factors are 2, 5
And 2*5=10

(viii) 3x y 2 3
= 3 × x × x × y × y × y
3 2
10x y = 2 × 5 × x × x × x × y × y
2 2
6x y z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y
And x × x × y × y = x y 2 2

Page : 220 , Block name : Exercise 14.1

Q2 Factorise the following expressions.

(i) 7x − 42
(ii) 6p – 12q
(iii) 7a + 14a2

(iv) −16z + 20z 3

(v) 20/ m + 30alm
2

(vi) 5x y − 15xy
2 2

(vii) 10a − 15b + 20c 2 2 2

(viii) −4a + 4ab − 4ca 2

(ix) x yz + xy z + xyz
2 2 2

(x) ax y + bxy + cxyz
2 2

Answer. (i) 7x = 7 × x
42 = 2 × 3 × 7

The common factor is 7
∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7(x − 6)

(ii) 6p = 2 × 3 × p
The common factors are 2 and 3.
12q = 2 × 2 × 3 × q

∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)

= 2 × 3[p − (2 × q)]

= 6(p − 2q)

(iii) 14a = 2 × 7 × a

2
∴ 7a + 14a = (7 × a × a) + (2 × 7 × a)

= 7 × a[a + 2] = 7a(a + 2)

(iv) 16z = 2 × 2 × 2 × 2 × z
3
20z = 2 × 2 × 5 × z × z × z

The common factors 2, 2, and z

Page 2 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
3
∴ −16z + 20z = −(2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)

= (2 × 2 × z)[−(2 × 2) + (5 × z × z)]
2
= 4z (−4 + 5z )

(v) 20/ m = 2 × 2 × 5 × / × l × m
2

30a/m = 2 × 3 × 5 × a × l × m

The common factors are 2,5,l and m
2
∴ 20/ m + 30a/m = (2 × 2 × 5 × I × l × m) + (2 × 3 × 5 × a × / × m)

= (2 × 5 × I × m)[(2 × l) + (3 × a)]

= 10 lm(2l + 3a)

(vi) 5x y = 5 × x × x × y
2

2
15xy = 3 × 5 × x × y × y

= 5 × x × y[x − (3 × y)]

= 5xy(x − 3y)

= 5xy(x − 3y)

(vii) 10a 2
= 2 × 5 × a × a
2
15b = 3 × 5 × b × b
2
20c = 2 × 2 × 5 × c × c

The common factor is 5.
2 2 2
10a − 15b + 20c = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)

= 5[(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]
2 2 2
= 5 (2a − 3b + 4c )

(viii) 4a 2
= 2 × 2 × a × a

4ab = 2 × 2 × a × b

4ca = 2 × 2 × c × a

The common factors are 2,2 and a
2
∴ −4a + 4ab − 4ca = −(2 × 2 × a × a) + (2 × 2 × a × b) − (2 × 2 × c × a)

= 2 × 2 × a[−(a) + b − c]

= 4a(−a + b − c)

(ix) x yz = x × x × y × z
2

2
xy z = x × y × y × z
2
xyz = x × y × z × z

The common factors are x, y, and z
2 2 2
∴ x yz + xy z + xyz = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)

= x × y × z[x + y + z]

= xyz(x + y + z)

(x) ax y = a × x × x × y
2

2
bxy = b × x × y × y

cxyz = c × x × y × z

The common factors are x and y
2 2
ax y + bxy + cxyz = (a × x × x × y) + (b × x × y × y) + (c × x × y × z)

= (x × y)[(a × x) + (b × y) + (c × z)]

= xy(ax + by + cz)

Page 3 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
Page : 220 , Block name : Exercise 14.1

Q3 Factorise
(i) x + xy + 8x + 8y
2

(ii) 15xy − 6x + 5y − 2
(iii) ax + bx − ay − by
(iv) 15pq + 15 + 9q + 25p
(v) z − 7 + 7xy − xyz

Answer. (i) x 2
+ xy + 8x + 8y = x × x + x × y + 8 × x + 8 × y

= x(x + y) + 8(x + y)

= (x + y)(x + 8)

(ii) 15xy − 6x + 5y − 2 = 3 × 5 × x × y − 3 × 2 × x + 5 × y − 2
= 3x(5y − 2) + 1(5y − 2)

= (5y − 2)(3x + 1)

(iii) ax + bx − ay − by = a × x + b × x − a × y − b × y
= x(a + b) − y(a + b)

= (a + b)(x − y)

(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15
= 3 × 5 × p × q + 3 × 3 × q + 5 × 5 × p + 3 × 5

= 3q(5p + 3) + 5(5p + 3)

(v) z − 7 + 7xy − xyz = z − x × y × z − 7 + 7 × x × y
= z(1 − xy) − 7(1 − xy)

= (1 − xy)(z − 7)

Page : 220 , Block name : Exercise 14.1

Exercise - 14.2

Q1 Factorise the following expressions
(i) a + 8a + 16
2

(ii)p − 10p + 25
x

(iii) 25m + 30m + 9
2

(iv) 49y + 84yz + 36z
4 2

(v) 4x − 8x + 4
2

(vi) 121b − 88bc + 16c
2 2

(vii) (l + m) − 4I m 2

(viii) a + 2a b + b
4 2 2 4

Answer. (i) a 2
+ 8a + 16 = (a)
2
+ 2 × a × 4 + (4)
2

2 2 2 2
= (a + 4) [(x + y) = x + 2xy + y ]

(ii) p
2
− 10p + 25 = (p)
2
− 2 × p × 5 + (5)
2

2 2 2 2
= (p − 5) [(a − b) = a − 2ab + b ]

Page 4 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
(iii) 25m 2
+ 30m + 9 = (5m)
2
+ 2 × 5m × 3 + (3)
2

2 2 2 2
= (5m + 3) [(a + b) = a + 2ab + b ]

(iv) 49y 2
+ 84yz + 36z
2
= (7y)
2
+ 2 × (7y) × (6z) + (6z)
2

2 2 2 2
= (7y + 6z) [(a + b) = a + 2ab + b ]

(v) 4x 2
− 8x + 4 = (2x)
2
− 2(2x)(2) + (2)
2

2 2 2 2
= (2x − 2) [(a − b) = a − 2ab + b ]

2 2
= [(2)(x − 1)] = 4(x − 1)

(vi) 121b 2
− 88bc + 16c
2
= (11b)
2
− 2(11b)(4c) + (4c)
2

2 2 2 2
= (11b − 4c) [(a − b) = a − 2ab + b ]

(v) 4x 2
− 8x + 4 = (2x)
2
− 2(2x)(2) + (2)
2

2 2 2 2
= (2x − 2) [(a − b) = a − 2ab + b ]

2 2
= [(2)(x − 1)] = 4(x − 1)

(vi) 121b 2
− 88bc + 16c
2
= (11b)
2
− 2(11b)(4c) + (4c)
2

2 2 2 2
= (11b − 4c) [(a − b) = a − 2ab + b ]

(vii) (I + m) 2
− 4lm = f
2
+ 2lm + m
2
− 4lm
2 2
= f − 2lm + m

2 2 2 2
= (I − m) [(a − b) = a − 2ab + b ]

2 2
(viii) a 4
+ 2a b
2 2
+ b
4
= (a )
2 2
+ 2 (a ) (b ) + (b )
2 2

2
2 2 2 2 2
= (a + b ) [(a + b) = a + 2ab + b ]

Page : 223 , Block name : Exercise 14.2

Q2 Factorise
(i) 4p = 9q
2 2

(ii) 63a − 112b
2 2

(iii) 49x − 36 2

(iv) 16x − 144x 3

(v) (l + m) − (l − m) 2 t

(vi) 9x y − 162 2

(vii) (x − 2xy + y ) − z
2 2 3

(viii) a 4
+ 2a b
2 2
+ b
4

Answer. (i) 4p 2
− 9q
2
= (2p)
2
− (3q)
2

2 2
= (2p + 3q)(2p − 3q) [a − b = (a − b)(a + b)]

(ii) 63a 2
− 112b
2
= 7 (9a
2
− 16b )
2

2 2
= 7 [(3a) − (4b) ]
2 2
= 7(3a + 4b)(3a − 4b) [a − b = (a − b)(a + b)]

Page 5 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
(iii) 49x 2
− 36 = (7x)
2
− (6)
2

2 2
= (7x − 6)(7x + 6) [a − b = (a − b)(a + b)]

(iv) 16x 5
− 144x
3
= 16x
3
(x
2
− 9)

3 2 2
= 16x [(x) − (3) ]

3 2 2
= 16x (x − 3)(x + 3) [a − b = (a − b)(a + b)]

(v) (I + m) 2
− (I − m)
2
= [(I + m) − (I − m)][(l + m) + (I − m)]
2 2
[ Using identity a − b = (a − b)(a + b)]

= (l + m − I + m)(/ + m + I − m)

= 2m × 2l

= 4ml

= 4lm

(vi) 9x y 2 2
− 16 = (3xy)
2
− (4)
2

2 2
= (3xy − 4)(3xy + 4) [a − b = (a − b)(a + b)]

(vii) (x 2
− 2xy + y ) − z
2 2
= (x − y)
2
− (z)
2
[(a − b)
2
= a
2 2
− 2ab + b ]

2 2
= (x − y − z)(x − y + z) [a − b = (a − b)(a + b)]

(viii) 25a 2
− 4b
2
+ 28bc − 49c
2
= 25a
2
− (4b
2
− 28bc + 49c )
2

2 2 2
(5a) − [(2b) − 2 × 2b × 7c + (7c) ]

= (5a)
2
− [(2b − 7c) ]
2
\
2 2 2
[ Using identity (a − b) = a − 2ab + b ]

= [5a + (2b − 7c)][5a − (2b − 7c)]
2 2
[ Using identity a − b = (a − b)(a + b)]

= (5a + 2b − 7c)(5a − 2b + 7c)

Page : 223 , Block name : Exercise 14.2

Q3 Factorise The Expressions.
(i) ax + bx2

(ii 7p + 21q
2 2

(iii) 2x + 2xy + 2xz
2 2 4

(iv) am + bm + bv + an
2 2 2 2

(v) (lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii) 5y − 20y − 8z + 2yz
2

(viii) 10ab + 4a + 5b + 2
(ix) 6xy − 4y + 6 − 9x

Answer. (i) ax 2
+ bx = a × x × x + b × x = x(ax + b)

(ii) 7p 2
+ 21q
2
= 7 × p × p + 3 × 7 × q × q = 7 (p
2
+ 3q )
2

(iii) 2x 3
+ 2xy
2
+ 2xz
2
= 2x (x
2
+ y
2
+ z )
2

(iv) am 2
+ bm
2
+ bn
2
+ an
2
= am
2
+ bm
2
+ an
2
+ bn
2

Page 6 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
2 2
m (a + b) + n (a + b)
2 2
= (a + b) (m + n )

(v) (I m + l) + m + 1 = I m + m + l + 1
= m(l + 1) + 1(I + 1)

= (I + 1)(m + 1)

(vi) y(y + z) + 9(y + z) = (y + z)(y + 9)

(vii) 5y 2
− 20y − 8z + 2yz = 5y
2
− 20y + 2yz − 8z

5y(y − 4) + 2z(y − 4))

(y − 4)(5y + 2z)

(viii) 10ab + 4a + 5b + 2 = 10ab + 5b + 4a + 2
= 5b(2a + 1) + 2(2a + 1)

= (2a + 1)(5b + 2)

(ix) 6xy − 4y + 6 − 9x = 6xy − 9x − 4y + 6
3x(2y − 3) − 2(2y − 3)

(2y − 3)(3x − 2)

Page : 223 , Block name : Exercise 14.2

Q4 Factorise

(i) a − b
4 4

(ii) p − 81
4

(iii) x − (y + z)
4 4

(iv)x − (x − z)
4 4

(v) a − 2a b + b
4 2 2 4

2 2
Answer. (i) a 4
− b
4
= (a )
2
− (b )
2

2 2 2 2
= (a − b ) (a + b )
2 2
= (a − b)(a + b) (a + b )

2
(ii) p4
− 81 = (p )
2
− (9)
2

2 2
= (p − 9) (p + 9)
2 2 2
= [(p) − (3) ] (p + 9)
2
= (p − 3)(p + 3) (p + 9)

2 2
(iii) x 4
− (y + z)
4
= (x )
2
− [(y + z) ]
2

2 2 2 2
= [x − (y + z) ] [x + (y + z) ]
2 2
= [x − (y + z)][x + (y + z)] [x + (y + z) ]
2 2
= (x − y − z)(x + y + z) [x + (y + z) ]

2 2
(iv) x 4
− (x − z)
4
= (x )
2
− [(x − z) ]
2

2 2 2 2
= [x − (x − z) ] [x + (x − z) ]

Page 7 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
2 2
= [x − (x − z)][x + (x − z)] [x + (x − z) ]

2 2 2
= z(2x − z) [x + x − 2xz + z ]

2 2
= z(2x − z) (2x − 2xz + z )

2 2
(v) a 4
− 2a b
2 2
+ b
4
= (a )
2 2 2
− 2 (a ) (b ) + (b )
2

2
2 2
= (a − b )
2
= [(a − b)(a + b)]
2 2
= (a − b) (a + b)

Page : 224 , Block name : Exercise 14.2

Q5 Factorise the following expressions

(i) p + 6p + 8
2

(ii) q − 10q + 21
2

(iii) p + 6p − 16
2

Answer. (i) p + 6p + 8 2

It can be observed that 8 = 4 × 2 and 4 + 2 = 6
2 2
∴ p + 6p + 8 = p + 2p + 4p + 8

= p(p + 2) + 4(p + 2)

= (p + 2)(p + 4)

(ii) q − 10q + 21
2

It can be observed that 21 = (−7) × (−3) and (−7) + (−3) = −10
2 2
∴ q − 10q + 21 = q − 7q − 3q + 21

= q(q − 7) − 3(q − 7)

= (q − 7)(q − 3)

(iii) p + 6p − 16
2

It can be observed that 16 = (−2) × 8 and 8 + (−2) = 6
2 2
p + 6p − 16 = p + 8p − 2p − 16

= p(p + 8) − 2(p + 8)

= (p + 8)(p − 2)

Page : 224 , Block name : Exercise 14.2

Exercise - 14.3

Q1 Carry out the following divisions

(i) 28x ÷ 56x 4

(ii) −36y ÷ 9y 3 2

(iii) 66pq r ÷ 11qr 2 3 2

(iv) 34x y z ÷ 51xy z
3 3 5 2 3

(v) 12a b ÷ (−6x b )
3 8 6 4

Answer. (i) 28x 4
= 2 × 2 × 7 × x × x × x × x

Page 8 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths
56x = 2 × 2 × 2 × 7 × x
3
4 2×2×7×x×x×x×x x 1 3
28x ÷ 56x = = = x
2×2×2×7×x 2 2

(ii) 36y 3
= 2 × 2 × 3 × 3 × y × y × y
2
9y = 3 × 3 × y × y
−2×2×3×3×y×y×y
3 2
−36y ÷ 9y = = −4y
3×3×y×y

(iii) 66pq r 2 3
= 2 × 3 × 11 × p × q × q × r × r × r
2
11qr = 11 × q × r × r
2 3 2 2×3×11×p×q×q×r×r×r
66pq r ÷ 11qr = = 6pqr
11×q×r×r

(iv) 34x y z 3 3 3
= 2 × 17 × x × x × x × y × y × y × z × z × z
2 3
51xy z = 3 × 17 × x × y × y × z × z × z
2×17×x×x×x×y×y×y×z×z×z
3 3 3 2 3
34x y z ÷ 51xy z =
3×17×x×y×z×z×z

2 2
= x y
3

(v) 12a b 8 8
= 2 × 2 × 3 × a
8
× b
8

6 4 6 4
6a b = 2 × 3 × a × b

8 8
8 8 6 4 2×2×3×a ×b 2 4
12a b ÷ (−6a b ) = 6 4
= −2a b
−2×3×a ×b

Page : 227 , Block name : Exercise 14.3

Q2 Divide the given polynomial by the given monomial.

(i) (5x 2
− 6x) ÷ 3x

(ii) (3y 8
− 4y
6
+ 5y ) ÷ y
4 4

(iii) 8 (x y z 3 2 2
+ x y z
2 3 2 2
+ x y z ) ÷ 4x y z
2 3 2 2 2

(iv) (x 3
+ 2x
2
+ 3x) ÷ 2x

(v) (p q 3 5 6
− p q ) ÷ p q
5 3 3

Answer. (i) 5x 2
− 6x = x(5x − 6)
x(5x−6) 1
2
(5x − 6x) ÷ 3x = = (5x − 6)
3x 3

(ii) 3y 8
− 4y
6
+ 5y
4
= y
4
(3y
4
− 4y
2
+ 5)
4 4 2
y (3y −4y +5)
8 6 4 4 4 2
(3y − 4y + 5y ) ÷ y = = 3y − 4y + 5
4
y

(iii) 8 (x y z 3 2 2
+ x y z
2 3 2 2 2
+ x y z ) = 8x y z (x + y + z)
3 2 2 2

2 2 2
8x y z (x+y+z)
3 2 2 2 3 2 2 2 3 2 2 2
8 (x y z + x y z + x y z ) ÷ 4x y z = 2 2 2
= 2(x + y + z)
4x y z

(iv) x 3
+ 2x
2
+ 3x = x (x
2
+ 2x + 3)
2
x(x +2x+3)
3 2 1 2
(x + 2x + 3x) ÷ 2x = = (x + 2x + 3)
2x 2

(v) p q 3 6
− p q
6 3
= p q
3 3
(q
3
− p )
3

3 3 3 3
p q (q −p )
3 6 6 3 3 3 3 3
(p q − p q ) ÷ p q = = q − p
3 3
p q

Page 9 of 16 Aglasem Schools

Book : Mathematics Ncert Solutions | Chapter-14 Maths

Page : 227 , Block name : Exercise 14.3

Q3 Work out the following divisions.

(i) (10x − 25) ÷ 5
(ii) (10x − 25) ÷ (2x − 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7)
(iv) 9x y (3z − 24) ÷ 27xy(z − 8)
2 2

(v) 96abc(3a − 12)(5b − 30) ÷ 144(a − 4)(b − 6)

5(2x−5)
Answer. (i) (10x − 25) ÷ 5 =
2×5×x−5×5
= = 2x − 5
5 5

5(2x−5)
(ii) (10x − 25) ÷ (2x − 5) =
2×5×x−5×5
= = 5
(2x−5) 2x−5

2×5×y[2×3×y+3×7]
(iii) 10y(6y + 21) ÷ 5(2y + 7) =
5(2y+7)

2×5×y×3(2y+7)
= = 6y
5(2y+7)

2 2
9x y [3×z−2×2×2×3]
(iv) 9x y (3z − 24) ÷ 27xy(z − 8) =
2 2

27xy(z−8)

xy×3(z−8)
= = xy
3(z−8)

(v) 96abc(3a − 12)(5b − 30) ÷ 144(a − 4)(b − 6)
96abc(3×a−3×4)(5×b−2×3×5)
=
144(a−4)(b−6)

2abc×3(a−4)×5(b−6)
= = 10abc
3(a−4)(b−6)

Page : 227 , Block name : Exercise 14.3

Q4 Divide as directed

(i) 5(2x + 1)(3x + 5) ÷ (2x + 1)
(ii) 26xy(x + 5)(y − 4) ÷ 13x(y − 4)
(iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
(iv) 20(y + 4) (y 2
+ 5y + 3) ÷ 5(y + 4)

(v) x(x + 1)(x + 2)(x + 3) ÷ (x + 1)

5(2x+1)(3x+1)
Answer. (i) 5(2x + 1)(3x + 5) ÷ (2x + 1) = (2x+1)
= 5(3x + 1)

2×13×xy(x+5)(y−4)
(ii) 26xy(x + 5)(y − 4) ÷ 13x(y − 4) = 13x(y−4)
= 2y(x + 5)

(iii) 52pqr(p + q)(q + r)(r + p) ÷ 104pq(q + r)(r + p)
2×2×13×p×q×r×(p+q)×(q+r)×(r+p)
=
2×2×2×13×p×q×(q+r)×(r+p)

1
= r(p + q)
2

(iv) 20(y + 4) (y 2
+ 5y + 3) = 2 × 2 × 5 × (y + 4) (y
2
+ 5y + 3)

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Book : Mathematics Ncert Solutions | Chapter-14 Maths
2
2×2×5×(y+4)×(y +5y+3)
2
20(y + 4) (y + 5y + 3) ÷ 5(y + 4) =
5×(y+4)

2
= 4 (y + 5y + 3)

x(x+1)(x+2)(x+3)
(v) x(x + 1)(x + 2)(x + 3) ÷ (x + 1) = x(x+1)

= (x + 2)(x + 3)

Page : 227 , Block name : Exercise 14.3

Q5 Factorise the expressions and divide them as directed

(i) (y 2
+ 7y + 10) ÷ (y + 5)

(ii) (m 2
− 14m − 32) ÷ (m + 2)

(iii) (5p 2
− 25p + 20) ÷ (p − 1)

(iv) 4yz (z 2
+ 6z − 16) ÷ 2y(z + 8)

(v) 5pq (p 2 2
− q ) ÷ 2p(p + q)

(vi) 12xy (9x 2 2
− 16y ) ÷ 4xy(3x + 4y)

(vii) 39y 3
(50y
2 2
− 98) ÷ 26y (5y + 7)

Answer. (i) (y 2
+ 7y + 10) = y
2
+ 2y + 5y + 10

= y(y + 2) + 5(y + 2)

= (y + 2)(y + 5)
(y+5)(y+2)
2
(y + 7y + 10) ÷ (y + 5) = = y + 2
(y+5)

(ii) m 2
− 14m − 32 = m
2
+ 2m − 16m − 32

= m(m + 2) − 16(m + 2)

= (m + 2)(m − 16)
(m+2)(m−16)
2
(m − 14m − 32) ÷ (m + 2) = = m − 16
(m+2)

(iii) 5p 2
− 25p + 20 = 5 (p
2
− 5p + 4)
2
= 5 [p − p − 4p + 4]

= 5[p(p − 1) − 4(p − 1)]

5(p − 1)(p − 4)
5(p−1)(p−4)
2
(5p − 25p + 20) ÷ (p − 1) = = 5(p − 4)
(p−1)

(iv) 4yz (z 2
+ 6z − 16) = 4yz [z
2
− 2z + 8z − 16]

= 4yz[z(z − 2) + 8(z − 2)]

= 4yz(z − 2)(z + 8)
4yz(z−2)(z+8)
2
4yz (z + 6z − 16) ÷ 2y(z + 8) = = 2z(z − 2)
2y(z+8)

(v) 5pq (p 2 2
− q ) = 5pq(p − q)(p + q)

5pq(p−q)(p+q) 5
2 2
5pq (p − q ) ÷ 2p(p + q) = = q(p − q)
2p(p+q) 2

(vi) 12xy (9x 2 2
− 16y ) = 12xy [(3x)
2 2
− (4y) ] = 12xy(3x − 4y)(3x + 4y)

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Book : Mathematics Ncert Solutions | Chapter-14 Maths
2 × 2 × 3 × x × y × (3x − 4y) × (3x + 4y)
2 2
12xy (9x − 16y ) ÷ 4xy(3x + 4y) =
2 × 2 × x × y × (3x + 4y)

= 3(3x − 4y)

= 3(3x − 4y)

(vii) 39y 3
(50y
2
− 98) = 3 × 13 × y × y × y × 2 [(25y
2
− 49)]

2 2
= 3 × 13 × 2 × y × y × y × [(5y) − (7) ]

= 3 × 13 × 2 × y × y × y(5y − 7)(5y + 7)
2
26y (5y + 7) = 2 × 13 × y × y × (5y + 7)
3 2 2
39y (50y − 98) ÷ 26y (5y + 7)

Page : 227 , Block name : Exercise 14.3

Exercise - 14.4

Q1 Find and correct the errors in the following mathematical statements.
4(x − 5) = 4x − 5

Answer. L.H.S = 4(x − 5) ≠ R.H.S
The correct statement is 4x - 20.

Page : 228 , Block name : Exercise 14.4

Q2 Find and correct the errors in the following mathematical statements.
2
x(3x + 2) = 3x + 2

Answer. L.H.S = x(3x + 2) ≠ R.H.S
The correct statement is 3x + 2x 2

Page : 228 , Block name : Exercise 14.4

Q3 Find and correct the errors in the following mathematical statements.
2x + 3y = 5xy

Answer. L.H.S = 2x + 3y ≠ R.H.S
The correct statement is 2x + 3y = 2x + 3y

Page : 228 , Block name : Exercise 14.4

Q4 Find and correct the errors in the following mathematical statements.
x + 2x + 3x = 5x

Answer. L.H.S = x + 2x + 3x = 1x + 2x + 3x = x(1 + 2 + 3) = 6x ≠ R.H.S
The correct statement is x + 2x + 3x = 6x

Page : 228 , Block name : Exercise 14.4

Q5 Find and correct the errors in the following mathematical statements.
5y + 2y + y − 7y = 0

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Book : Mathematics Ncert Solutions | Chapter-14 Maths

Answer. L.H.S = 5y + 2y + y − 7y = 8y − 7y = y ≠ R.H.S
The correct statement is 5y + 2y + y − 7y = y

Page : 228 , Block name : Exercise 14.4

Q6 Find and correct the errors in the following mathematical statements.
2
3x + 2x = 5x

Answer. L.H.S = 3x + 2x = 5x ≠ R.H.S
The correct statement is 3x + 2x = 5x

Page : 228 , Block name : Exercise 14.4

Q7 Find and correct the errors in the following mathematical statements.
2 2
(2x) + 4(2x) + 7 = 2x + 8x + 7

Answer. L.H.S = (2x) 2
+ 4(2x) + 7 = 4x
2
+ 8x + 7 ≠ R.H.S
The correct statement is (2x) 2
+ 4(2x) + 7 = 4x
2
+ 8x + 7

Page : 228 , Block name : Exercise 14.4

Q8 Find and correct the errors in the following mathematical statements.
2 2
(2x) + 4(2x) + 7 = 2x + 8x + 7

Answer. L.H.S = (2x) 2
+ 5x = 4x
2
+ 5x ≠ R.H.S
The correct statement is (2x) 2
+ 5x = 4x
2
+ 5x

Page : 228 , Block name : Exercise 14.4

Q9 Find and correct the errors in the following mathematical statements. (3x + 2) 2
= 3x
2
+ 6x + 4

Answer. L.H.S = (3x + 2) 2
= (3x)
2
+ 2(3x)(2) + (2)
2
[(a + b)
2
= a
2 2
+ 2ab + b ] = 9x
2
+ 12x + 4

R.H.S
The correct statement is (3x + 2) 2
= 9x
2
+ 12x + 4

Page : 228 , Block name : Exercise 14.4

Q10 Find and correct the errors in the following mathematical statements. Substituting x=-3

(a) x 2
+ 5x + 4 gives (−3) 2
+ 5(−3) + 4 = 9 + 2 + 4 = 15

(b) x 2
− 5x + 4 gives (−3) 2
− 5(−3) + 4 = 9 − 15 + 4 = −2

(c) x 2
+ 5x gives (−3) 2
+ 5(−3) = −9 − 15 = −24

Answer. (a) For x = −3
2 2
x + 5x + 4 = (−3) + 5(−3) + 4 = 9 − 15 + 4 = 13 − 15 = −2

(b) For x = −3
2 2
x − 5x + 4 = (−3) − 5(−3) + 4 = 9 + 15 + 4 = 28

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Book : Mathematics Ncert Solutions | Chapter-14 Maths

(c) For x = −3
2 2
x + 5x = (−3) + 5(−3) = 9 − 15 = −6

Page : 229 , Block name : Exercise 14.4

Q11 Find and correct the errors in the following mathematical statements.
2 2
(y − 3) = y − 9

Answer. L.H.S = (y − 3) 2
= (y)
2
− 2(y)(3) + (3)
2
[(a − b)
2
= a
2
− 2ab + b ]
2

2
= y − 6y + 9 ≠ R. H . S

Page : 229 , Block name : Exercise 14.4

Q12 Find and correct the error in the statement (z + 5) 2
= z
2
+ 25

Answer. L.H.S. = (z + 5) 2
= (z)
2
+ 2(z)(5) + (5)
2
[(a + b)
2
= a
2
+ 2ab + b ]
2

= z
2
R.H.S
+ 10z + 25 ≠

The correct statement is (z + 5) 2
= z
2
+ 10z + 25

Page : 229 , Block name : Exercise 14.4

Q13 Find and correct the error in the statement
2 2
(2a + 3b)(a − b) = 2a − 3b

Answer. L.H.S.
2 2 2 2
= (2a + 3b)(a − b) = 2a × a + 3b × a − 2a × b − 3b × b = 2a + 3ab − 2ab − 3b = 2a + ab − 3b ≠

R.H.S
The correct statement is (2a + 3b)(a − b) = 2a 2
+ ab − 3b
2

Page : 229 , Block name : Exercise 14.4

Q14 Find and correct the error in the statement
2
(a + 4)(a + 2) = a + 8

Answer. L.H.S = (a − 4)(a − 2) = (a) 2
+ [(−4) + (−2)](a) + (−4)(−2) = a
2
+ 6a + 8 ≠ R. H . S

The correct statement is (a + 4)(a + 2) = a 2
+ 6a + 8

Page : 229 , Block name : Exercise 14.4

Q15 Find and correct the error in the statement (a − 4)(a − 2) = a 2
− 8

Answer. L.H.S. = (a − 4)(a − 2) = (a) 2
+ [(−4) + (−2)](a) + (−4)(−2) = a
2
− 6a + 8 ≠ R. H . S

The correct statement is (a − 4)(a − 2) = a 2
− 6a + 8

Page : 229 , Block name : Exercise 14.4

Q16 Find and correct the error in the statement
2
3x
= 0
3x2

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Book : Mathematics Ncert Solutions | Chapter-14 Maths
Answer. Not equal to 0.
2
3x
2
= 1
3x

Page : 229 , Block name : Exercise 14.4

Q17 Find and correct the error in the statement
2
3x +1

2
− 1 + 1 = 2
3x

2 2

Answer.
3x +1 3x 1 1
2
= 2
+ 2
= 1 + 2
3x 3x 3x 3x

And not equal to 2.

Page : 229 , Block name : Exercise 14.4

Q18 Find and correct the error in the statement
3x 1
=
3x+2 2

Answer. 3x+2
3x
=
3x

3x+2

And is not equal to 1

2

Page : 229 , Block name : Exercise 14.4

Q19 Find and correct the error in the statement
3 1
=
4x+3 4x

Answer.
3 3
=
4x+3 4x+3

And is not equal to .
1

4x

Page : 229 , Block name : Exercise 14.4

Q20 Find and correct the error in the statement
4x+5
= 5
4x

Answer.
4x+5 4x 5 5
= + = 1 + and not equal to 5
4x 4x 4x 4x

Page : 229 , Block name : Exercise 14.4

Q21 Find and correct the error in the statement
7x+5
− 7x
5

Answer.
7x+5 7x 5 7x
= + = + 1 and not equal to 7x
5 5 5 5

Page : 229 , Block name : Exercise 14.4

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Book : Mathematics Ncert Solutions | Chapter-14 Maths

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