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NCERT Solutions for Class 8 Maths Chapter 14 – Factorisation


Exercise 14.1 : Solutions of Questions on Page Number : 220


Q1 : Find the common factors of the terms
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16x3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z

Answer :
(i) 12x = 2 x 2 x 3 x x
36 = 2 x 2 x 3 x 3
The common factors are 2, 2, 3.
And, 2 x 2 x 3 = 12
(ii) 2y = 2 x y
22xy = 2 x 11 x x x y
The common factors are 2, y.
And, 2 x y = 2y
(iii) 14pq = 2 x 7 x p x q
28p2q2 = 2 x 2 x 7 x p x p x q x q
The common factors are 2, 7, p, q.
And, 2 x 7 x p x q = 14pq
(iv) 2x = 2 x x
3x2 = 3 x x x x
4 = 2 x 2
The common factor is 1.
(v) 6abc = 2 x 3 x a x b x c
24ab2 = 2 x 2 x 2 x 3 x a x b x b
12a2b = 2 x 2 x 3 x a x a x b
The common factors are 2, 3, a, b.
And, 2 x 3 x a x b = 6ab
(vi) 16x3 = 2 x 2 x 2 x 2 x x x x x x
-4x2 = -1 x 2 x 2 x x x x
32x = 2 x 2 x 2 x 2 x 2 x x
The common factors are 2, 2, x.
And, 2 x 2 x x = 4x
(vii) 10pq = 2 x 5 x p x q
20qr = 2 x 2 x 5 x q x r
30rp = 2 x 3 x 5 x r x p
The common factors are 2, 5.
And, 2 x 5 = 10
(viii) 3x2y3 = 3 x x x x x y x y x y
10x3y2 = 2 x 5 x x x x x x x y x y
6x2y2z = 2 x 3 x x x x x y x y x z
The common factors are x, x, y, y.
And,
x x x x y x y = x2y2


Q2 :Factorise the following expressions
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a2 + 14a
(iv) -16z + 20z3
(v) 20l2m + 30 alm
(vi) 5x2y – 15xy2
(vii) 10a2 – 15b2 + 20c2
(viii) -4a2 + 4ab – 4 ca
(ix) x2yz + xy2z + xyz2
(x) ax2y + bxy2 + cxyz

Answer :
(i) 7x = 7 x x
42 = 2 x 3 x 7
The common factor is 7.
∴ 7x – 42 = (7 x x) – (2 x 3 x 7) = 7 (x – 6)
(ii) 6p = 2 x 3 x p
12q = 2 x 2 x 3 x q
The common factors are 2 and 3.
∴ 6p – 12q = (2 x 3 x p) – (2 x 2 x 3 x q)
= 2 x 3 [p – (2 x q)]
= 6 (p – 2q)
(iii) 7a2 = 7 x a x a
14a = 2 x 7 x a
The common factors are 7 and a.
∴ 7a2 + 14a = (7 x a x a) + (2 x 7 x a)
= 7 x a [a + 2] = 7a (a + 2)
(iv) 16z = 2 x 2 x 2 x 2 x z
20z3 = 2 x 2 x 5 x z x z x z
The common factors are 2, 2, and z.
∴ -16z + 20z3 = – (2 x 2 x 2 x 2 x z) + (2 x 2 x 5 x z x z x z)
= (2 x 2 x z) [- (2 x 2) + (5 x z x z)]
= 4z (- 4 + 5z2)
(v) 20l2m = 2 x 2 x 5 x l x l x m
30alm = 2 x 3 x 5 x a x l x m
The common factors are 2, 5, l, and m.
∴ 20l2m + 30alm = (2 x 2 x 5 x l x l x m) + (2 x 3 x 5 x a x l x m)
= (2 x 5 x l x m) [(2 x l) + (3 x a)]
= 10lm (2l + 3a)
(vi) 5x2y = 5 x x x x x y
15xy2 = 3 x 5 x x x y x y
The common factors are 5, x, and y.
∴ 5x2y – 15xy2 = (5 x x x x x y) – (3 x 5 x x x y x y)
= 5 x x x y [x – (3 x y)]
= 5xy (x – 3y)
(vii) 10a2 = 2 x 5 x a x a
15b2 = 3 x 5 x b x b
20c2 = 2 x 2 x 5 x c x c
The common factor is 5.
10a2 – 15b2 + 20c2 = (2 x 5 x a x a) – (3 x 5 x b x b) + (2 x 2 x 5 x c x c)
= 5 [(2 x a x a) – (3 x b x b) + (2 x 2 x c x c)]
= 5 (2a2 – 3b2 + 4c2)
(viii) 4a2 = 2 x 2 x a x a
4ab = 2 x 2 x a x b
4ca = 2 x 2 x c x a
The common factors are 2, 2, and a.
∴ -4a2 + 4ab – 4ca = – (2 x 2 x a x a) + (2 x 2 x a x b) – (2 x 2 x c x a)
= 2 x 2 x a [- (a) + bc]
= 4a (-a + bc)
(ix) x2yz = x x x x y x z
xy2z = x x y x y x z
xyz2 = x x y x z x z
The common factors are x, y, and z.
x2yz + xy2z + xyz2 = (x x x x y x z) + (x x y x y x z) + (x x y x z x z)
= x x y x z [x + y + z]
= xyz (x + y + z)
(x) ax2y = a x x x x x y
bxy2 = b x x x y x y
cxyz = c x x x y x z


Q3 : Factorise
(i) x2 + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bxayby
(iv) 15pq + 15 + 9q + 25p
(v) z – 7 + 7xyxyz
Answer :
(i) x2 + xy + 8x + 8y = x x x + x x y + 8 x x + 8 x y
= x (x + y) + 8 (x + y)
= (x + y) (x + 8)
(ii) 15xy – 6x + 5y – 2 = 3 x 5 x x x y – 3 x 2 x x + 5 x y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)
(iii) ax + bxayby = a x x + b x xa x yb x y
= x (a + b) – y (a + b)
= (a + b) (xy)
(iv) 15pq + 15 + 9q + 25p = 15pq + 9q + 25p + 15
= 3 x 5 x p x q + 3 x 3 x q + 5 x 5 x p + 3 x 5
= 3q (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)
(v) z – 7 + 7xyxyz = zx x y x z – 7 + 7 x x x y
= z (1 – xy) – 7 (1 – xy)
= (1 – xy) (z – 7)


Exercise 14.2 : Solutions of Questions on Page Number : 223


Q1 :Factorise the following expressions.
(i) a2 + 8a + 16
(ii) p2 – 10p + 25
(iii) 25m2 + 30m + 9
(iv) 49y2 + 84yz + 36z2
(v) 4x2 – 8x + 4
(vi) 121b2 – 88bc + 16c2
(vii) (l + m)2 – 4lm (Hint: Expand (l + m)2 first)
(viii) a4 + 2a2b2 + b4

Answer :
(i) a2 + 8a + 16 = (a)2 + 2 x a x 4 + (4)2
= (a + 4)2 [(x + y)2 = x2 + 2xy + y2]
(ii) p2 – 10p + 25 = (p)2 – 2 x p x 5 + (5)2
= (p – 5)2 [(ab)2 = a2 – 2ab + b2]
(iii) 25m2 + 30m + 9 = (5m)2 + 2 x 5m x 3 + (3)2
= (5m + 3)2 [(a + b)2 = a2 + 2ab + b2]
(iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 x (7y) x (6z) + (6z)2
= (7y + 6z)2 [(a + b)2 = a2 + 2ab + b2]
(v) 4x2 – 8x + 4 = (2x)2 – 2 (2x) (2) + (2)2
= (2x – 2)2 [(ab)2 = a2 – 2ab + b2]
= [(2) (x – 1)]2 = 4(x – 1)2
(vi) 121b2 – 88bc + 16c2 = (11b)2 – 2 (11b) (4c) + (4c)2
= (11b – 4c)2 [(ab)2 = a2 – 2ab + b2]
(vii) (l + m)2 – 4lm = l2 + 2lm + m2 – 4lm
= l2 – 2lm + m2
= (lm)2 [(ab)2 = a2 – 2ab + b2]
(viii) a4 + 2a2b2 + b4 = (a2)2 + 2 (a2) (b2) + (b2)2
= (a2 + b2)2 [(a + b)2 = a2 + 2ab + b2]


Q2 :Factorise
(i) 4p2 – 9q2
(ii) 63a2 – 112b2
(iii) 49x2 – 36
(iv) 16x5 – 144x3
(v) (l + m)2 – (lm)2
(vi) 9x2y2 – 16
(vii) (x2 – 2xy + y2) – z2
(viii) 25a2 – 4b2 + 28bc – 49c2

Answer :
(i) 4p2 – 9q2 = (2p)2 – (3q)2
= (2p + 3q) (2p – 3q) [a2b2 = (ab) (a + b)]
(ii) 63a2 – 112b2 = 7(9a2 – 16b2)
= 7[(3a)2 – (4b)2]
= 7(3a + 4b) (3a – 4b) [a2b2 = (ab) (a + b)]
(iii) 49x2 – 36 = (7x)2 – (6)2
= (7x – 6) (7x + 6) [a2b2 = (ab) (a + b)]
(iv) 16x5 – 144x3 = 16x3(x2 – 9)
= 16 x3 [(x)2 – (3)2]
= 16 x3(x – 3) (x + 3) [a2b2 = (ab) (a + b)]
(v) (l + m)2 – (lm)2 = [(l + m) – (lm)] [(l + m) + (lm)]
[Using identity a2b2 = (ab) (a + b)]
= (l + ml + m) (l + m + lm)
= 2m x 2l
= 4ml
= 4lm
(vi) 9x2y2 – 16 = (3xy)2 – (4)2
= (3xy – 4) (3xy + 4) [a2b2 = (ab) (a + b)]
(vii) (x2 – 2xy + y2) – z2 = (xy)2 – (z)2 [(ab)2 = a2 – 2ab + b2]
= (xyz) (xy + z) [a2b2 = (ab) (a + b)]
(viii) 25a2 – 4b2 + 28bc – 49c2 = 25a2 – (4b2 – 28bc + 49c2)
= (5a)2 – [(2b)2 – 2 x 2b x 7c + (7c)2]
= (5a)2 – [(2b – 7c)2]
[Using identity (ab)2 = a2 – 2ab + b2]
= [5a + (2b – 7c)] [5a – (2b – 7c)]
[Using identity a2b2 = (ab) (a + b)]
= (5a + 2b – 7c) (5a – 2b + 7c)


Q3 :Factorise the expressions
(i) ax2 + bx
(ii) 7p2 + 21q2
(iii) 2x3 + 2xy2 + 2xz2
(iv) am2 + bm2 + bn2 + an2
(v) (lm + l) + m + 1
(vi) y(y + z) + 9(y + z)
(vii) 5y2 – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Answer :
(i) ax2 + bx = a x x x x + b x x = x(ax + b)
(ii) 7p2 + 21q2 = 7 x p x p + 3 x 7 x q x q = 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2 = 2x(x2 + y2 + z2)
(iv) am2 + bm2 + bn2 + an2 = am2 + bm2 + an2 + bn2
= m2(a + b) + n2(a + b)
= (a + b) (m2 + n2)
(v) (lm + l) + m + 1 = lm + m + l + 1
= m(l + 1) + 1(l + 1)
= (l + l) (m + 1)
(vi) y (y + z) + 9 (y + z) = (y + z) (y + 9)
(vii) 5y2 – 20y – 8z + 2yz = 5y2 – 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)


Q4 : Factorise
(i) a4b4
(ii) p4 – 81
(iii) x4 – (y + z)4
(iv) x4 – (xz)4
(v) a4 – 2a2b2 + b4

Answer :
(i) a4b4 = (a2)2 – (b2)2
= (a2b2) (a2 + b2)
= (ab) (a + b) (a2 + b2)
(ii) p4 – 81 = (p2)2 – (9)2
= (p2 – 9) (p2 + 9)
= [(p)2 – (3)2] (p2 + 9)
= (p – 3) (p + 3) (p2 + 9)
(iii) x4 – (y + z)4 = (x2)2 – [(y +z)2]2
= [x2 – (y + z)2] [x2 + (y + z)2]
= [x – (y + z)][ x + (y + z)] [x2 + (y + z)2]
= (xyz) (x + y + z) [x2 + (y + z)2]
(iv) x4 – (xz)4 = (x2)2 – [(xz)2]2
= [x2 – (xz)2] [x2 + (xz)2]
= [x – (xz)] [x + (xz)] [x2 + (xz)2]
= z(2xz) [x2 + x2 – 2xz + z2]
= z(2xz) (2x2 – 2xz + z2)
(v) a4 – 2a2b2 + b4 = (a2)2 – 2 (a2) (b2) + (b2)2
= (a2b2)2
= [(ab) (a + b)]2
= (ab)2 (a + b)2


Q5 : Factorise the following expressions
(i) p2 + 6p + 8
(ii) q2 – 10q + 21
(iii) p2 + 6p – 16

Answer :
(i) p2 + 6p + 8
It can be observed that, 8 = 4 x 2 and 4 + 2 = 6
p2 + 6p + 8 = p2 + 2p + 4p + 8
= p(p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q2 – 10q + 21
It can be observed that, 21 = (-7) x (-3) and (-7) + (-3) = – 10
q2 – 10q + 21 = q2 – 7q – 3q + 21
= q(q – 7) – 3(q – 7)
= (q – 7) (q – 3)
(iii) p2 + 6p – 16
It can be observed that, 16 = (-2) x 8 and 8 + (-2) = 6
p2 + 6p – 16 = p2 + 8p – 2p – 16
= p(p + 8) – 2(p + 8)
= (p + 8) (p – 2)


Exercise 14.3 : Solutions of Questions on Page Number : 227


Q1 :Carry out the following divisions.
(i) 28x4 ÷ 56x
(ii) -36y3 ÷  9y2
(iii) 66pq2r3 ÷ 11qr2
(iv) 34x3y3z3 ÷ 51xy2z3
(v) 12a8b8 ÷  (-6a6b4)
Answer :
(i) 28x4 = 2 × 2 × 7 × x × x × x × x
56x = 2 × 2 × 2 × 7 × x

(ii) 36y3 = 2 × 2 × 3 × 3 × y × y × y
9y2 = 3 × 3 × y × y

(iii) 66 pq2 r3 = 2 × 3 × 11 × p × q × q × r × r × r
11qr2 = 11 × q × r × r

(iv) 34 x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z
51 xy2z3 = 3 ×17 × x × y × y ×z × z × z

(v) 12a8b8 = 2 × 2 × 3 × a8 × b8
6a6b4 = 2 × 3 × a6 × b4
= – 2a2b4


Q2 : Divide the given polynomial by the given monomial.
(i) (5x2 – 6x) ÷3x
(ii) (3y8 – 4y6 + 5y4) ÷ y4
(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
(iv) (x3 + 2x2 + 3x) ÷ 2x
(v) (p3q6p6q3p3q3
Answer :
(i) 5x2 – 6x = x(5x – 6)

(ii) 3y8 – 4y6 + 5y4 = y4(3y4 – 4y2 + 5)

(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)

(iv) x3 + 2x2 + 3x = x(x2 + 2x + 3)

(v) p3q6p6q3 = p3q3(q3p3)


Q3 : Work out the following divisions.
(i) (10x – 25) ÷ 5
(ii) (10x – 25) ÷ (2x – 5)
(iii) 10y(6y + 21) ÷ 5(2y + 7)
(iv) 9x2y2(3z – 24)÷ 27xy(z – 8)
(v) 96abc(3a – 12)(5b – 30) ÷ 144(a – 4) (b – 6)
Answer :
(i)
(ii)
(iii)

(iv)

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


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) (y2 + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1) (x + 2) (x + 3) ÷ x(x + 1)
Answer :
(i) = 5(3x + 1)
(ii) = 2y (x + 5)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)

(iv) 20(y + 4) (y2 + 5y + 3) = 2 × 2 × 5 × (y + 4) (y2 + 5y + 3)

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


Q5 :Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10)÷ (y + 5)
(ii) (m2 – 14m – 32) ÷ (m + 2)
(iii) (5p2 – 25p + 20) ÷ (p – 1)
(iv) 4yz(z2 + 6z – 16)÷2y(z + 8)
(v) 5pq(p2q2)÷ 2p(p + q)
(vi) 12xy(9x2 – 16y2) ÷4xy(3x + 4y)
(vii) 39y3(50y2– 98) ÷ 26y2(5y+ 7)

Answer :
(i) (y2 + 7y + 10) = y2 + 2y + 5y + 10
= y (y + 2) + 5 (y + 2)
= (y + 2) (y + 5)

(ii) m2 – 14m – 32 = m2 + 2m – 16m – 32
= m (m + 2) – 16 (m + 2)
= (m + 2) (m – 16)

(iii) 5p2 – 25p + 20 = 5(p2 – 5p + 4)
= 5[p2p – 4p + 4]
= 5[p(p – 1) – 4(p – 1)]
= 5(p – 1) (p – 4)

(iv) 4yz(z2 + 6z – 16) = 4yz [z2 – 2z + 8z – 16]
= 4yz [z(z – 2) + 8(z – 2)]
= 4yz(z – 2) (z + 8)

(v) 5pq(p2q2) = 5pq (pq) (p + q)

(vi) 12xy(9x2 – 16y2) = 12xy[(3x)2 – (4y)2] = 12xy(3x – 4y) (3x + 4y)

(vii) 39y3(50y2 – 98) = 3 × 13 × y × y × y × 2[(25y2 – 49)]
= 3 × 13 × 2 × y × y × y × [(5y)2 – (7)2]
= 3 × 13 × 2 × y × y × y (5y – 7) (5y + 7)
26y2(5y + 7) = 2 × 13 × y × y × (5y + 7)
39y3(50y2 – 98) ÷26y2 (5y + 7)


Exercise 14.4 : Solutions of Questions on Page Number : 228


Q1 : Find and correct the errors in the statement: 4(x – 5) = 4x – 5

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


Q2 : Find and correct the errors in the statement: x(3x + 2) = 3x2 + 2

Answer :
L.H.S. = x(3x + 2) = x x 3x + x x 2 = 3x2 + 2x ≠ R.H.S.
The correct statement is x(3x + 2) = 3x2 + 2x


Q3 : Find and correct the errors in the statement: 2x + 3y = 5xy

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


Q4 : Find and correct the errors in the statement: 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


Q5 : Find and correct the errors in the statement: 5y + 2y + y – 7y = 0

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


Q6 : Find and correct the errors in the statement: 3x + 2x = 5x2

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


Q7 :  Find and correct the errors in the statement: (2x)2 + 4(2x) + 7 = 2x2 + 8x + 7

Answer :
L.H.S = (2x)2 + 4(2x) + 7 = 4x2 + 8x + 7 ≠ R.H.S
The correct statement is (2x)2 + 4(2x) + 7 = 4x2 + 8x + 7


Q8 : Find and correct the errors in the statement: (2x)2 + 5x = 4x + 5x = 9x

Answer :
L.H.S = (2x)2 + 5x = 4x2 + 5x ≠ R.H.S.
The correct statement is (2x)2 + 5x = 4x2 + 5x


Q9 : Find and correct the errors in the statement: (3x + 2)2 = 3x2 + 6x + 4

Answer :
L.H.S. = (3x + 2)2 = (3x)2 + 2(3x)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2]
= 9x2 + 12x + 4 ≠ R.H.S
The correct statement is (3x + 2)2 = 9x2 + 12x + 4


Q10 : Find and correct the errors in the statement: (y – 3)2 = y2 – 9

Answer :
L.H.S = (y – 3)2 = (y)2 – 2(y)(3) + (3)2 [(ab)2 = a2 – 2ab + b2]
= y2 – 6y + 9 ≠ R.H.S
The correct statement is (y – 3)2 = y2 – 6y + 9


Q11 :Find and correct the errors in the statement: (z + 5)2 = z2 + 25

Answer :
L.H.S = (z + 5)2 = (z)2 + 2(z)(5) + (5)2 [(a + b)2 = a2 + 2ab + b2]
= z2 + 10z + 25 ≠ R.H.S
The correct statement is (z + 5)2 = z2 + 10z + 25


Q12 :Find and correct the errors in the statement: (2a + 3b) (ab) = 2a2 – 3b2

Answer :
L.H.S. = (2a + 3b) (ab) = 2a x a + 3b x a – 2a x b – 3b x b
= 2a2 + 3ab – 2ab – 3b2 = 2a2 + ab – 3b2 ≠ R.H.S.
The correct statement is (2a + 3b) (ab) = 2a2 + ab – 3b2


Q13 : Find and correct the errors in the statement: (a + 4) (a + 2) = a2 + 8

Answer :
L.H.S. = (a + 4) (a + 2) = (a)2 + (4 + 2) (a) + 4 x 2
= a2 + 6a + 8 ≠ R.H.S
The correct statement is (a + 4) (a + 2) = a2 + 6a + 8


Q14 : Find and correct the errors in the statement: (a – 4) (a – 2) = a2 – 8

Answer :
L.H.S. = (a – 4) (a – 2) = (a)2 + [(- 4) + (- 2)] (a) + (- 4) (- 2)
= a2 – 6a + 8 ≠ R.H.S.
The correct statement is (a – 4) (a – 2) = a2 – 6a + 8



Q16 : Find and correct the errors in the statement:
Answer :

The correct statement is


Q17 : Find and correct the errors in the statement:
Answer :
L.H.S =
The correct statement is


Q18 :Find and correct the errors in the statement:
Answer :
L.H.S. =
The correct statement is 


Q19 :Find and correct the errors in the statement:
Answer :
L.H.S. =≠ R.H.S.
The correct statement is 


Q20 :Find and correct the errors in the statement: 
Answer :
L.H.S. = 
The correct statement is 


Q21 :Find and correct the errors in the statement: 
Answer :
L.H.S. = 
The correct statement is