Exercise -14.1
Question 1- 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 × 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
28p2q2 = 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
3x2 = 3 × x × x
4 = 2 × 2
The common factor is 1.
(v) 6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
The common factors are 2, 3, a, b.
And, 2 × 3 × a × b = 6ab
(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x
−4x2 = −1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors are 2, 2, x.
And, 2 × 2 × x = 4x
(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) 3x2y3 = 3 × x × x × y × y × y
10x3y2 = 2 × 5 × x × x × x × y × y
6x2y2z = 2 × 3 × x × x × y × y × z
The common factors are x, x, y, y.
And,
x × x × y × y = x2y2
Question 2-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
42 = 2 × 3 × 7
The common factor is 7.
∴ 7x − 42 = (7 × x) − (2 × 3 × 7) = 7 (x − 6)
(ii) 6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors are 2 and 3.
∴ 6p − 12q = (2 × 3 × p) − (2 × 2 × 3 × q)
= 2 × 3 [p − (2 × q)]
= 6 (p − 2q)
(iii) 7a2 = 7 × a × a
14a = 2 × 7 × a
The common factors are 7 and a.
∴ 7a2 + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a [a + 2] = 7a (a + 2)
(iv) 16z = 2 × 2 × 2 × 2 × z
20z3 = 2 × 2 × 5 × z × z × z
The common factors are 2, 2, and z.
∴ −16z + 20z3 = − (2 × 2 × 2 × 2 × z) + (2 × 2 × 5 × z × z × z)
= (2 × 2 × z) [− (2 × 2) + (5 × z × z)]
= 4z (− 4 + 5z2)
(v) 20l2m = 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
The common factors are 2, 5, l, and m.
∴ 20l2m + 30alm = (2 × 2 × 5 × l × l × m) + (2 × 3 × 5 × a × l × m)
= (2 × 5 × l × m) [(2 × l) + (3 × a)]
= 10lm (2l + 3a)
(vi) 5x2y = 5 × x × x × y
15xy2 = 3 × 5 × x × y × y
The common factors are 5, x, and y.
∴ 5x2y − 15xy2 = (5 × x × x × y) − (3 × 5 × x × y × y)
= 5 × x × y [x − (3 × y)]
= 5xy (x − 3y)
(vii) 10a2 = 2 × 5 × a × a
15b2 = 3 × 5 × b × b
20c2 = 2 × 2 × 5 × c × c
The common factor is 5.
10a2 − 15b2 + 20c2 = (2 × 5 × a × a) − (3 × 5 × b × b) + (2 × 2 × 5 × c × c)
= 5 [(2 × a × a) − (3 × b × b) + (2 × 2 × c × c)]
= 5 (2a2 − 3b2 + 4c2)
(viii) 4a2 = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca = 2 × 2 × c × a
The common factors are 2, 2, and a.
∴ −4a2 + 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) x2yz = x × x × y × z
xy2z = x × y × y × z
xyz2 = x × y × z × z
The common factors are x, y, and z.
∴ x2yz + xy2z + xyz2 = (x × x × y × z) + (x × y × y × z) + (x × y × z × z)
= x × y × z [x + y + z]
= xyz (x + y + z)
(x) ax2y = a × x × x × y
bxy2 = b × x × y × y
cxyz = c × x × y × z
The common factors are x and y.
ax2y + bxy2 + 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)
Question 3- Factorise
(i) x2 + xy + 8x + 8y
(ii) 15xy − 6x + 5y − 2
(iii) ax + bx − ay − by
(iv) 15pq + 15 + 9q + 25p
(v) z − 7 + 7xy − xyz
Answer- (i) x2 + 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)
= (5p + 3) (3q + 5)
(v) z − 7 + 7xy − xyz = z − x × y × z − 7 + 7 × x × y
= z (1 − xy) − 7 (1 − xy)
= (1 − xy) (z − 7)
Exercise -14.2
Question 1- 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 × a × 4 + (4)2
= (a + 4)2 [(x + y)2 = x2 + 2xy + y2]
(ii) p2 − 10p + 25 = (p)2 − 2 × p × 5 + (5)2
= (p − 5)2 [(a − b)2 = a2 − 2ab + b2]
(iii) 25m2 + 30m + 9 = (5m)2 + 2 × 5m × 3 + (3)2
= (5m + 3)2 [(a + b)2 = a2 + 2ab + b2]
(iv) 49y2 + 84yz + 36z2 = (7y)2 + 2 × (7y) × (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 [(a − b)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 [(a − b)2 = a2 − 2ab + b2]
(vii) (l + m)2 − 4lm = l2 + 2lm + m2 − 4lm
= l2 − 2lm + m2
= (l − m)2 [(a − b)2 = a2 − 2ab + b2]
(viii) a4 + 2a2b2 + b4 = (a2)2 + 2 (a2) (b2) + (b2)2
= (a2 + b2)2 [(a + b)2 = a2 + 2ab + b2]
Question- 2 Factorise
(i) 4p2 − 9q2
(ii) 63a2 − 112b2
(iii) 49x2 − 36
(iv) 16x5 − 144x3
(v) (l + m)2 − (l − m)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) [a2 − b2 = (a − b) (a + b)]
(ii) 63a2 − 112b2 = 7(9a2 − 16b2)
= 7[(3a)2 − (4b)2]
= 7(3a + 4b) (3a − 4b) [a2 − b2 = (a − b) (a + b)]
(iii) 49x2 − 36 = (7x)2 − (6)2
= (7x − 6) (7x + 6) [a2 − b2 = (a − b) (a + b)]
(iv) 16x5 − 144x3 = 16x3(x2 − 9)
= 16 x3 [(x)2 − (3)2]
= 16 x3(x − 3) (x + 3) [a2 − b2 = (a − b) (a + b)]
(v) (l + m)2 − (l − m)2 = [(l + m) − (l − m)] [(l + m) + (l − m)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (l + m − l + m) (l + m + l − m)
= 2m × 2l
= 4ml
= 4lm
(vi) 9x2y2 − 16 = (3xy)2 − (4)2
= (3xy − 4) (3xy + 4) [a2 − b2 = (a − b) (a + b)]
(vii) (x2 − 2xy + y2) − z2 = (x − y)2 − (z)2 [(a − b)2 = a2 − 2ab + b2]
= (x − y − z) (x − y + z) [a2 − b2 = (a − b) (a + b)]
(viii) 25a2 − 4b2 + 28bc − 49c2 = 25a2 − (4b2 − 28bc + 49c2)
= (5a)2 − [(2b)2 − 2 × 2b × 7c + (7c)2]
= (5a)2 − [(2b − 7c)2]
[Using identity (a − b)2 = a2 − 2ab + b2]
= [5a + (2b − 7c)] [5a − (2b − 7c)]
[Using identity a2 − b2 = (a − b) (a + b)]
= (5a + 2b − 7c) (5a − 2b + 7c)
Question 3- 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 + b × x = x(ax + b)
(ii) 7p2 + 21q2 = 7 × p × p + 3 × 7 × q × 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)
Question 4- Factorise
(i) a4 − b4
(ii) p4 − 81
(iii) x4 − (y + z)4
(iv) x4 − (x − z)4
(v) a4 − 2a2b2 + b4
Answer - (i) a4 − b4 = (a2)2 − (b2)2
= (a2 − b2) (a2 + b2)
= (a − b) (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]
= (x − y − z) (x + y + z) [x2 + (y + z)2]
(iv) x4 − (x − z)4 = (x2)2 − [(x − z)2]2
= [x2 − (x − z)2] [x2 + (x − z)2]
= [x − (x − z)] [x + (x − z)] [x2 + (x − z)2]
= z(2x − z) [x2 + x2 − 2xz + z2]
= z(2x − z) (2x2 − 2xz + z2)
(v) a4 − 2a2b2 + b4 = (a2)2 − 2 (a2) (b2) + (b2)2
= (a2 − b2)2
= [(a − b) (a + b)]2
= (a − b)2 (a + b)2
Question 5- 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 × 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) × (−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) × 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
Question 1- 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
Question 2- 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) (p3q6 − p6q3) ÷ p3q3
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) p3q6− p6q3 = p3q3(q3 − p3)
Question 3- 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)
Question 4- 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)
= 5(3x + 1)
(ii) 
= 2y (x + 5)
= 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)
Question 5- 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(p2 − q2) ÷ 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[p2 − p − 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(p2 − q2) = 5pq (p − q) (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
Question 1- Find and correct the errors in the statement: 4(x − 5) = 4x − 5
Answer - L.H.S. = 4(x − 5) = 4 × x − 4 × 5 = 4x − 20 ≠ R.H.S.
The correct statement is 4(x − 5) = 4x − 20
Question 2- Find and correct the errors in the statement: x(3x + 2) = 3x2 + 2
Answer - L.H.S. = x(3x + 2) = x × 3x + x × 2 = 3x2 + 2x ≠ R.H.S.
The correct statement is x(3x + 2) = 3x2 + 2x
Question 3- 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
Question 4- 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
Question 5- 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
Question 6- 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
Question 7- 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
Question 8- 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
Question 9- 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
Question 10- Find and correct the errors in the following mathematical statement. Substituting x = −3 in
(a) x2 + 5x + 4 gives (−3)2 + 5 (−3) + 4 = 9 + 2 + 4 = 15
(b) x2 − 5x + 4 gives (−3)2 − 5 (−3) + 4 = 9 − 15 + 4 = −2
(c) x2 + 5x gives (−3)2 + 5 (−3) = −9 − 15 = −24
Answer - (a) For x = −3,
x2 + 5x + 4 = (−3)2 + 5 (−3) + 4 = 9 − 15 + 4 = 13 − 15 = −2
(b) For x = −3,
x2 − 5x + 4 = (−3)2 − 5 (−3) + 4 = 9 + 15 + 4 = 28
(c) For x = −3,
x2 + 5x = (−3)2 + 5(−3) = 9 − 15 = −6
Question 11- 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 [(a − b)2 = a2 − 2ab + b2]
= y2 − 6y + 9 ≠ R.H.S
The correct statement is (y − 3)2 = y2 − 6y + 9
Question 12- 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
Question 13- Find and correct the errors in the statement: (2a + 3b) (a − b) = 2a2 − 3b2
Answer - L.H.S. = (2a + 3b) (a − b) = 2a × a + 3b × a − 2a × b − 3b × b
= 2a2 + 3ab − 2ab − 3b2 = 2a2 + ab − 3b2 ≠ R.H.S.
The correct statement is (2a + 3b) (a − b) = 2a2 + ab − 3b2
Question 14- 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
Question 15- 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
Question 16- Find and correct the errors in the statement: 
Answer - L.H.S = 
The correct statement is 
Question 17- Find and correct the errors in the statement: 
Answer -
The correct statement is 
Question 18- Find and correct the errors in the statement: 
Answer - L.H.S =
The correct statement is 
Question 19- Find and correct the errors in the statement: 
Answer - L.H.S. =
≠ R.H.S.
≠ R.H.S.
The correct statement is 
Question 20- Find and correct the errors in the statement: 
Answer - L.H.S. =
The correct statement is 
Question 21- Find and correct the errors in the statement: 
Answer - L.H.S. =
The correct statement is
The correct statement is
its very helpful..........thanks !!!!!!!
ReplyDeletethanks very much, quite helpful in exams
ReplyDeleteWow! Thank you so much for the solution sir.
ReplyDeleteTHANKS sir for the solution with full method.I think i will secure good marks in exam by practicing your solutions. THANK YOU VERY MUCH.........
ReplyDeleteThanks for help sir.
ReplyDelete