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Question 1. Find the common factors of the given terms in each.
• (i) 8x, 24
• (ii) 3a, 2lab
• (iii) 7xy, 35x2y3
• (iv) 4m2, 6m2, 8m3
• (v) 15p, 20qr, 25rp
• (vi) 4x2, 6xy, 8y2x
• (vii) 12 x2y, 18xy2

Solution:
8x = 2 × 2 × 2 × x
24 = 8 × 3 = 2 × 2 × 2 × 3
∴ Common factors of 8x, 24 = 2, 4, 8.

ii) 3a, 2lab
3a = 3 × a
21ab = 7 × 3 × a × b
∴ Common factors of 3a, 21ab = 3, a, 3a.

iii) 7xy, 35x2y3
7xy = 7 × x × y
35x2y3 = 7 × 5 × x × x × y × y × y
∴ Common factors of 7xy, 35x2y3
= 7, x, y, 7x, 7y, xy, 7xy.

iv) 4m2, 6m2, 8m3
4m2 = 2 × 2 × m × m
6m2 = 2 × 3 × m × m
8m3 = 2 × 2 × 2 × m × m × m
∴ Common factors of 4m2, 6m2, 8m3
= 2, m, m2, 2m, 2m2.

v) 15p, 20qr, 25rp
15p = 3 × 5 × p
20qr = 4 × 5 × q × r
25rp = 5 × 5 × r × p
∴ Common factors of 15p, 20qr, 25rp = 5.

vi) 4x2, 6xy, 8y2x
4x2= 2 × 2 × x × x
6xy = 2 × 3 × x × y
8y2x = 2 × 2 × 2 × y × y × x
∴ Common factors of 4x2, 6xy, 8xy2= 2, x, 2x.

vii) 12x2y, 18xy2
12x22y = 2 × 2 × 3 × x × x × y
18xy2 = 3 × 3 × 2 × x × y × y
∴ Common factors of 12x2y, 18xy2
= 2,3, x, y, 6, xy, 6x, 6y, 2x, 2y, 3x, 3y, 6xy.

Question 2. Factorise the following expressions
• (i) 5x2- 25xy
• (ii) 9a2- 6ax
• (iii) 7p2+ 49pq
• (iv) 36 a2b - 60 a2bc
• (v) 3a2bc + 6ab2c + 9abc2
• (vi) 4p2+ 5pq - 6pq2
• (vii) ut + at2

Solution:
(i) 5x2 - 25xy
= 5 x × x × - 5 × 5 × x × y
= 5 × x [x - 5 × y]
= 5x [x - 5y]

ii) 9a2- 6ax
= 3 × 3 × a × a - 2 × 3 × a × x
= 3a [3a - 2x]

iii) 7p2+ 49pq
= 7 × p × p +7 × 7 × p × q
= 7p[p + 7q]

iv) 36a2b - 60a2bc
= 2 × 2 × 3 × 3 × a × a × b - 2 × 2 × 3 × 5 × a × a × b × c
= 2 × 2 × 3 × a × a × b[3 - 5c]
= 12a2b [3 - 5c]

v) 3a2bc + 6ab2c + 9abc2
= 3 × a × a × b × c + 3 × 2 × a × b × b × c + 3 × 3 × a × b × c × c
= 3abc [a + 2b + 3c]

vi) 4p2+ 5pq - 6pq2
= 2 × 2 × p × p + 5 × p × q - 2 × 3 × p × q × q
= p [4p + 5q - 6q2]

vii) ut + at2
= u × t + a × t × t = t [u + at]

Question 3. Factorise the following:
• (i) 3ax – 6xy + 8by - 4bx
• (ii) x3+ 2x2+ 5x + 10
• (iii) m2 - mn + 4m - 4n
• (iv) a3- a2b2 - ab + b3
• (v) p2q - pr2 - pq + r2

Solution:
i) 3ax - 6xy + 8by - 4ab
= (3ax - 6xy) -(4ab - 8by)
= (3 × a × x - 2 × 3 × x × y)
- (4 ×a × b - 4 × 2 × b × y)
= 3x(a - 2y) - 4b(a - 2y)
= (a - 2y)(3x - 4b)

ii) x3 + 2x2 + 5x + 10
= (x3 + 2x2) + (5x +10)
= (x2 × x + 2 × x2) + (5 × x + 5 × 2)
= x2(x + 2) + 5(x + 2)
= (x + 2) (x2 + 5)

iii) m2 - mn + 4m - 4n
= (m2 - mn) + (4m - 4n)
= (m × m - m × n) + (4 × m - 4 × n)
= m(m - n) + 4(m - n)
= (m - n) (m + 4)

iv) a3 - a2b2 - ab + b3
= (a3 - a2b2) - (ab - b3)
= (a2 × a - a2 × b2) - (a × b - b × b2)
= a2(a - b2) - b(a - b2)
= (a - b2) (a2- b)

v) p21 - pr2 - pq + r2
= (p2q - pr2) - (pq - r2)
= (p × p × q - p × r × r) - (pq - r2)
= p(pq - r2) - (pq - r2) × 1
= (p - 1) (pq - r2)

Question 1. Factorise the following expression
• i) a2 + 10a +25
• ii) l2 - 16l + 64
• iii) 36x2 + 96xy + 64y2
• iv) 25x2 + 9y2 - 30xy
• v) 25m2 - 40mn + 1 6n2
• vi) 81x2 - 198 xy + 12ly2
• vii) (x+y)2 - 4xy
• (Hint : first expand ( x + y)2)
• viii) l4 + 4l2m2 + 4m4

Solution:
i) a2 + 10a +25
= (a)2 + 2 × a × 5 + (5)2
It is in the form of a2 + 2ab + b2
a2 + 2ab + b2= (a + b)2
∴ a2 + 10a + 25 = (a + 5)2 = (a + 5) (a + 5)

ii) l2 - 16l + 64
l2 - 16l + 64
= (l)2 - 2 × l × 8 + (8)2
It is in the form of a2 - 2ab + b2
a2 - 2ab + b2 = (a - b)2
∴ l2 - 16l + 64 = (l - 8)2 = (l - 8) (l - 8)

iii) 36x2 + 96xy + 64y2
36x2 + 96xy + 64y2
= (6x)2 + 2 × 6x × 8y + (8y)2
It is in the form of a2 + 2ab + b2
a2 + 2ab + b2 = (a + b)2
∴ 36x2 + 96xy + 64y2
= (6x + 8y)2 = (6x + 8y) (6x + 8y)

iv) 25x2+ 9y2 - 30xy
25x2 + 9y2 - 30xy
= (5x)2 + (3y)2 - 2 × 5x × 3y
It is in the form of a2 + b2 - 2ab
a2 + b2 - 2ab = (a - b)2
∴ 25x2 + 9y2 - 30xy
= (5x - 3y)2 = (5x - 3y) (5x - 3y)

v) 25m2 - 40mn + 1 6n2
25m2 - 40mn + 16n2
= (5m)2 - 2 × 5m × 4n + (4n)2
It is in the form of a2 - 2ab + b2
a2 - 2ab + b2 = (a - b)2
∴ 25m2 - 40mn + 16n2
= (5m - 4n)2
= (5m - 4n) (5m - 4n)

vi) 81x2 - 198 xy + 12ly2
81x2 - 198xy + 121y2
= (9x)2 - 2 × 9x × 11y + (11y)2
It is in the form of a22
a2 - 2ab + b2 = (a - b)2
∴ 81x2 - 198xy + 121y2
= (9x - 11y)2 - (9x - 11y) (9x - 11y)

vii) (x+y)2 - 4xy
(Hint : first expand ( x + y)2)
= (x + y)2- 4xy
= x2 + y2 + 2xy - 4xy
= x2 + y2 - 2xy = (x - y)2 = (x - y)(x - y)

viii) l4+ 4l2m2+ 4m4
l4+ 4l2m2+ 4m4
= (l2)2+ 2 × l2× 2m2+ (2m2)2
It is in the form of a2+ 2ab + b2
a2+ 2ab + b2= (a - b)2
∴ l4+ 4l2m2+ 4m4
= (l2+ 2m2)2= (l2+ 2m2) (l2+ 2m2)

Question 2.
Factorise the following

• i) x2 - 36
• ii) 49x2 - 25y2
• iii) m2 - 121
• iv) 81 - 64x2
• v) x2y2 - 64
• vi) 6x2 - 54
• vii) x2 - 81
• viii) 2x -32 x5
• ix) 81x4 - 121x2
• x) (p2 - 2pq + q2)-r2
• xi) (x+y)2 - (x-y)2

Solution:
i) x2 - 36
x2 - 36
⇒ (x)2 - (6)2 is in the form of a2 - b2
a2 - b2 = (a + b) (a - b)
∴ x2 - 36 = (x + 6) (x - 6)

ii) 49x2 - 25y2
= (7x)2 - (5y)2
= (7x + 5y) (7x - 5y)

iii) m2 - 121
m2-121
= (m)2 - (11)2
= (m + 11) (m - 11)

iv) 81 - 64x2
81 - 64x2
= (9)2 - (8x)2
= (9 + 8x) (9 - 8x)

v) x2y2 - 64
= (xy)2 - (8)2
= (xy + 8)(xy - 8)

vi) 6x2 - 54
6x2 - 54
= 6x2 - 6 x 9 ‘
= 6(x2 - 9)
= 6[(x)2 - (3)2]
= 6(x + 3) (x - 3)

vii) x2 - 81
x2 - 81
= x2 - 92
= (x + 9 )(x - 9)

viii) 2x - 32 x5
2x - 32 x5
= 2x - 2x x 16x4
= 2 x (1 - 16x4)
= 2x [12) - (4x2)2]
= 2x (1 + 4x2) (1 - 4x2)
= 2x (1 + 4x2) [(15 - (2x)2]
= 2x (1 + 4x2) (1 + 2x) (1 - 2x)

ix) 81x4 - 121x2
81x4 - 121x2
- x2 (812 - 121)
= x2[(9x)2 - (11)2]
= x2(9x + 11) (9x -11)

x) (p2 - 2pq + q2)-r2
(p2 - 2pq + q2) - r2
= (p - q)2 - (r)2 [∵ p2 - 2pq + q2= (p - q)2]
= (p - q + r) (p - q - r)

xi) (x + y)2 - (x - y)2
(x + y)2 - (x - y)2
It is in the form of a2 - b2
a = x + y, b = x- y
∴ a2 - b2=(a + b)(a-b)
= (x + y + x - y) [x + y- (x - y)]
= 2x [x + y-x + y]
= 2x x 2y = 4xy

Question 3. Factorise the expressions

(i) lx2 + mx
(ii) 7y2 + 35Z2
(iii) 3x4 + 6x3y + 9x2Z
(iv) x2 - ax - bx + ab
(v) 3ax - 6ay - 8by + 4bx
(vi) mn + m + n + 1
(vii) 6ab - b2 + 12ac - 2bc
(viii) p2q - pr2 - pq + r2
(ix) x (y + z) -5 (y + z)

Solution

(i) lx2 + mx
lx2 + mx
= l × x × x + m × x = x(lx + m)

(ii) 7y2+ 35z2
7y2+ 35z2
= 7 × y2+ 7 × 5 × z2
= 7(y2+ 5z2)

(iii) 3x4+ 6x3y + 9x2Z
3x4+ 6x3y + 9x2Z
= 3 × x2 × x2 + 3 × 2 × x × x2× y + 3 × 3 × x2× z
= 3x2(x2 + 2xy + 3z)

(iv) x2 - ax - bx + ab
x2- ax - bx + ab
= (x2 - ax) - (bx - ab)
= x(x - a) - b(x - a)
= (x - a) (x - b)

(v) 3ax - 6ay - 8by + 4bx
3ax - 6ay - 8by + 4bx
= (3ax - 6ay) + (4bx - 8by)
= 3a (x - 2y) + 4b (x - 2y)
= (x - 2y) (3a + 4b)

(vi) mn + m + n + 1
mn + m + n + 1
= (mn + m) + (n + 1)
= m (n + 1) + (n + 1)
= (n + 1) (m + 1)

(vii) 6ab - b2 + 12ac - 2bc
6ab - b2+ 12ac - 2bc
= (6ab - b2) + (12ac - 2bc)
= (6 × a× b - b × b) + (6 × 2 × a × c - 2 × b × c)
= b [6a - b] + 2c [6a - b]
= (6a - b) (b + 2c)

(viii) p2q - pr2 - pq + r2
p2q - pr2 - pq + r2
= (p2q - pr2) - (pq - r2)
= (p × p × q - p × r × r) - (pq - r2)
= P(pq - r2) - (pq - r2) × 1
= (pq - r2)(p - 1)

(ix) x (y + z) -5 (y + z)
= x(y + z) - 5(y + z)
= (y + z) (x - 5)

Question 4. Factorise the following
• (i) x4 - y4
• (ii) a4 - (b + c)4
• (iii) l2 - (m - n)2
• (iv) 49x2 - 16/25
• (v) x4 - 2x2y2 + y4
• (vi) 4 (a + b)2 - 9 (a - b)2

Solution:
= (x2)2 - (y2)2is in the form of a2- b2
a2 - b2= (a + b) (a - b)
x4 - y4= (x2+ y2)(x2- y2)
= (x2+ y2)(x + y)(x - y)

(ii) a4- (b + c)4
a4- (b + c)4
= (a2)2- [(b + c)2]2
= [a2+ (b + c)2] [a2- (b + c)2] ,
= [a2+ (b + c)2] (a + b + c) [a - (b + c)]
= [a2+ (b + c)2] (a + b + c) (a - b - c)

(iii) l2- (m - n)2
l2 - (m - n)2
= (l)2 - (m - n)2
= [l + m - n] [l - (m - n)]
= [l + m -n] [l - m + n]

(iv) 49x2 - 16/25
= (7x)2 - (4/5)2
= (7x+ (45) (7x - (4/5)

(v) x4- 2x2 y 2 + y4
= (x2)2 - 2x2y2+ (y2)2
It is in the form of a2- 2ab + b2
a2 - 2ab + b2= (a - b)2
∴ x4- 2x2y2+ y4= (x2- y2)2
= [(x)2- (y)2]2
= [(x + y) (x - y)]2
= (x + y)2(x - y)2
[∵ (ab)m= am. bn]

(vi) 4 (a + b)2- 9 (a - b)2
4 (a + b)2- 9 (a - b)2
= [2(a + b)]2- [3(a - b)]2
= [2(a + b) + 3(a- b)] [2(a + b)-3(a- b)]
= (2a + 2b + 3a - 3b) (2a + 2b - 3a + 3b)
= (5a - b) (5b - a)

Question 5.
Factorise the following expressions

• (i) a2+ 10a + 24
• (ii) x2+9x + 18
• (iii) p2 - 10q + 21
• (iv) x2 - 4x - 32

Solution:
(i) a2+ 10a + 24
a2+ 10a + 24 .
= a2+ 6a + 4a + 24
= a x a + 6a + 4a + 6 x 4
= a(a + 6) + 4(a + 6)
= (a + 6) (a + 4) (or)
a2+ 10a + 24

∴ a2+ 10a + 24 = (a + 6) (a + 4)

(ii) x2+ 9x + 18
x2+ 9x + 18
= (x + 3) (x + 6)

∴ x2+ 9x + 18 = (x + 3) (x + 6)

(iii) p2- 10q + 21
p2 - 10p + 21
= (P - 7) (p - 3)

∴ p2- 10p + 21 = (p - 7)(p - 3)

(iv) x2 - 4x - 32
x2 - 4x - 32
= (x - 8) (x + 4)

∴ x2 – 4x – 32 = (x – 8) (x + 4)

Question 6. The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.

Solution:
Perimeter of a triangle
= AB + BC + CA = 48
⇒ c + a + b = 48

The solutions of Harmeet, Rosy are wrong.

∴ Srikar had done it correctly.
⇒ 21 + a + b = 48
⇒ a + b = 48 - 21 = 27
∴ The lengths of a, b should be 10, 17
∴ a + b > c [the sum of any two sides of a triangle is greater than the 3rd side]
∴ 10 + 17 > 2
27 > 21 (T).
∴ The length of the shortest side is 10 cm.

Question 7. Find the values of ‘m’ for which x2+ 3xy + x + my - in has two linear factors in x and y, with integer coefficients.

Solution:
Given equation is x2+ 3xy + x + my - m ……….(1)
Let the two linear equations in x and y be (x + 3y + a) and (x + 0y + b).
Then (x + 3y + a) (x + 0y + b)
= x2+ 0xy + bx + 3xy + 0y2+ 3by + ax + 0y + ab
= x2+ bx + ax + 3xy + 3by + ab ………….. (2)
Comparing equation (2) with (1),
x2+ 3xy + x + my - m
= x2+ (a + b)x + 3xy + 3by + ab
Equating the like terms on both sides,
ab = - m ………….. (3)
(a + b)x = x ⇒ a + b = 1 ……………. (4)
3by = my ⇒ 3b = m ⇒ b = m/3
Substitute ‘b’ value in equation (4),
a =1?m/3=3?m/3
ab = -m
[ ∵ from (3)]
put a & b value then ,

Question 1. Carry out the following divisions
• (i) 48a3 by 6a
• (ii) 14xby 42x3
• (iii) 72a3b4c5 by 8ab2c3
• (iv) 11xy2z3 by 55xyz
• (v) -54l4m3n2 by 9l2m2n2

Solution:
(i) 48a3 by 6a
48a3 ÷ 6a
=6×8×a×a2/ 6×a
= 8a2

(ii) 14x3by 42x3
= 14x3÷ 42x3

(iii) 72a3b4c5 by 8ab2c3

(iv) 11xy2z3by 55xyz
11xy2z3 ÷ 55xyz

(v) -54l4m3n2 by 9l2m2n2
-54l4m3n2÷ 9l2m2n2

= -6l2m

Question 2.
Divide the given polynomial by the given monomial

• (i) (3x2- 2x) ÷ x
• (ii) (5a3b - 7ab3) ÷ ab
• (iii) (25x5 - 15x4) ÷ 5x3
• (iv) (4l5 - 6l4+ 8l3) ÷ 2l2
• (v) 15 (a3b2c2- a2b3c2+ a2b2c3) ÷ 3abc
• (vi) 3p3- 9p2q - 6pq2) ÷ (-3p)
• (vii) (2/3 a2b2c2+4/3 a b2 c3) ÷12abc

Solution:
(i) (3x2 - 2x) ÷ x

(ii) (5a3b – 7ab3) ÷ ab

(iii) (25x5- 15x4) ÷ 5x3

= 5x2 - 3x (or) x(5x - 3)

(iv) (4l5 - 6l4+ 8l3) ÷ 2l2

= 2l2- 3l2 + 4l = l(2l2- 3l + 4)

(v) 15 (a3b2c2 - a2 b3c2+ a2b2c3) ÷ 3abc

= 5[a x abc - b x abc + c x abc ]
= 5abc [a - b + c]

(vi) 3p3 - 9p2q - 6pq2) ÷ (-3p)

= -[p2 - 3pq - 2q2]
= 22+ 3pq - p2

Question 3. Workout the following divisions:
• (i) (49x -63) ÷ 7
• (ii) 12x (8x - 20,) ÷ 4(2x - 5)
• (iii) 11a3b3(7c - 35) ÷ 3a2b2(c - 5)
• (iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)
• (v) 36(x + 4)(x2+ 7x + 10) ÷ 9(x + 4)
• (vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

Solution:
(i) (49x -63) ÷ 7

(ii) 12x (8x - 20,) ÷ 4(2x - 5)

(iii) 11a3b3(7c - 35) ÷ 3a2b2(c - 5)

(iv) 54lmn (l + m) (m + n) (n + l) ÷ 8 lmn (l + m) (n +l)

(v) 36(x + 4)(x2 + 7x + 10) ÷ 9(x + 4)

4 ( x2+ 7x + 10)
= 4 ( x2+ 5x + 2x + 10)
= 4 [x( x + 5) +2(x + 5)]
= 4( x + 5) (x + 2)

(vi) a(a+1)(a+2)(a + 3) ÷ a(a + 3)

= ( a + 1)(a + 2)

Question 4. Factorize the expressions and divide them as directed:
• (i) (x2+ 7x + 12) ÷ (x + 3)
• (ii) (x2- 8x + 12) ÷ (x - 6)
• (iii) (p2+ 5p + 4,) (p + l)
• (iv) 15ab(a2- 7a + 10) ÷ 3b(a - 2)
• (v) 151m (2p2- 2q2) ÷ 3l(p + q)
• (vi) 26z3(32z2 - 18,) ÷ 13z2(4z - 3)

Solution:
(i) (x2+ 7x + 12) ÷ (x + 3)
(x2+ 7x + 12) ÷ (x + 3)
x2+ 7x + 12 = x2+ 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)

(ii) (x2 - 8x + 12) ÷ (x - 6)
(x2 - 8x + 12)÷ (x - 6)
x2 - 8x + 12 = x2 - 6x - 2x + 12
= x(x - 6) - 2(x - 6)
= (x - 6) (x - 2)
∴ (x2 - 8x + 12) 4 (x - 6)
= (x?6)(x?2)/(x?6) = x - 2

(iii) (p2 + 5p + 4,) (p + 1)
p2 + 5p + 4 = p2 + p + 4p + 4
= p(p + 1) + 4(p + 1)
= (p + 1) (p + 4)
(p2 + 5p + 4) ÷ (p + 1)
= (p+1)(p+4)/(p+1) = p + 4

(iv) 15ab(a2 - 7a + 10) ÷ 3b(a - 2)
15ab (a2 - 7a + 10) ÷ 3b (a - 2)
15ab (a2 - 7a + 10) = 15ab (a2 - 5a - 2a + 10)
= 15ab [(a2 - 2a) - (5a -10)]
= 15ab [a(a - 2) - 5(a - 2)]
= 15ab(a - 2)(a - 5)
∴ 15ab (a2 - 7a + 10) ÷ 3b (a - 2)

(v) 151m (2p2 - 2q2) ÷ 3l(p + q)
15lm (2p2 - 2q2) ÷ 3l (p + q)
15lm (2p2 - 2q2) = 15lm x 2(p2 - q2)
= 30lm (p + q) (p - q)
∴ 15lm(2p2 - 2q2) ÷ 3l(p + q)

(vi) 26z3(32z2 - 18,)÷ 13z2 (4z - 3)
26z3(32z2 - 18) ÷ 13z2 (4z - 3)
26z3(32z2 - 18) = 26z3 (2 x 16z2 - 2 x 9)
= 26z3 x 2 [16z3 - 9]
= 52z3 [(4z)3 - (3)3]
= 52z3 (4z + 3) (4z - 3)
∴ 26z3 (32z2 - 18) ÷ 13z2 (4z - 3)

Question 1. Find the errors and correct the following mathematical sentences
• (i) 3(x – 9) = 3x - 9
• (ii) x(3x+2) = 3x2+ 2
• (iii) 2x+3x = 5x2
• (iv) 2x + x + 3x = sx
• (v) 4p + 3p + 2p + p - 9p = 0
• (vi) 3x + 2y = 6xy
• (vii) (3x)2+ 4x +7 = 3x2+ 4x +7
• (viii) (2x)2+ 5x = 4x + 5x = 9x
• (ix) (2a + 3)2= 2a2+ 6a +9
• (x) Substitute x -3 in
• (a) x2+ 7x + 12 (- 3)2+ 7(-3) + 12 = 9 + 4 + 12 = 25
• (b) x2- 5x + 6(-3)2- 5(-3) + 69 - 15 + 6 = 0
• (c) x2+5x = (-3)2+ 5(3) + 6 = -9 - 15 = -24
• (xi) (x - 4)2= x2- 16
• (xii) (x + 7)2= x2+49
• (xiii) (3a + 4b)(a - b)= 3a2- 4a2
• (xiv) (x + 4) (x + 2) = x2+ 8
• (xv) (x - 4) (x - 2) = x2- 8
• (xvi) 5x3÷ 5 x3= 0
• (xvii) 2x3+ 1 ÷ 2x3= 1
• (xviii) 3x + 2 ÷ 3x = 2/3x
• (xix) 3x + 5 ÷ 3 = 5
• (xx)4x+3/3= x + 1

Solution:
(i) 3(x - 9) = 3x - 9
3(x - 9) = 3x - 9
⇒ 3x - 3 x 9 = 3x - 9
⇒ 3x - 27 = 3x - 9
⇒ - 27 ≠ - 9
∴ The given sentence is wrong. Correct sentence is 3(x - 9) = 3x - 27.

(ii) x(3x+2) = 3x2 + 2
x(3x + 2) = 3x2 + 2
⇒ x × 3x + x × 2 = 3x2 + 2
⇒ 3x2 + 2x ≠ 3x2 + 2
∴ The given sentence is wrong.
Correct sentence is x(3x + 2) = 3x2 + 2x.

(iii) 2x+3x = 5x2
2x + 3x = 5x2
⇒ 5x = 5x2
⇒ x ≠ x2
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.

(iv) 2x + x + 3x = 5x
2x + x + 3x = 5x
⇒ 6x = 5x
⇒ 6 ≠ 5
∴ The given sentence is wrong. Correct sentence is 2x + 3x = 5x.
(v) 4p + 3p + 2p + p - 9p = 0
4p + 3p + 2p + p - 9p = 0
⇒ 10p - 9p = 0
⇒ p = 0
It is not possible
∴ The given sentence is wrong. Correct sentence is
4p + 3p + 2p + p - 9p - p = 0

(vi) 3x + 2y = 6xy
3x + 2y = 6xy
a + b ≠ ab
∴ The given sentence is wrong.
Correct sentence is 3x x 2y = 6xy.
(vii) (3x)2 + 4x +7 = 3x2 + 4x +7
(3x)2 + 4x +7 = 3x2 + 4x +7
⇒ (3x)2 = 3x2
⇒ 9x2 = 3x2
⇒ 9 = 3
It is not possible
∴ The given sentence is wrong. Correct sentence is
(3x)2+ 4x + 7 = 9x2 + 4x + 7.

(viii) (2x)2 + 5x = 4x + 5x = 9x
(2x)2 + 5x = 4x + 5x = 9x
⇒ 4x2 + 5x = 4x + 5x
⇒ 4x2 = 4x
⇒ x2 = x
⇒ x ≠ √x
∴ The given sentence is wrong. Correct sentence is (2x)2 + 5x = 4x2 + 5x.
(ix) (2a + 3)2 = 2a2 + 6a +9
(2a + 3)2 = 2a2 + 6a +9
⇒ (2a)2 + 2 × 2a × 3 + 32 = 2a2 + 6a + 9
⇒ 4a2 + 12a + 9 = 2a2+ 6a + 9
⇒ 4a2 - 2a2 = 6a - 12a
⇒ 2a2 = - 6a
⇒ 2a ≠ 6
∴The given sentence is wrong.
Correct sentence is
(2a + 3)2 = 4a2 + 12a + 9.

(x) Substitute x -3 in
(a) x2 + 7x + 12 (- 3)2 + 7(-3) + 12 = 9 + 4 + 12 = 25
x2 + 7x + 12 = (- 3)2 + 7 (- 3) + 12
= 9 - 21 + 12
= 21 - 21
= 0 25 (False)

(b) x2 - 5x + 6(-3)2 - 5(-3) + 69 - 15 + 6 = 0
x2 - 5x + 6 = (-3)2 - 5 (- 3) + 6
= 9 + 15 + 6
= 30 ≠ 0 (False)
(c) x2 +5x = (-3)2 + 5(3) + 6 = -9 - 15 = -24
x2 + 5x = (- 3)2 + 5 (- 3)
= 9 - 15 = - 6 ≠ 24 (False)
(xi) (x - 4)2 = x2 - 16
(x - 4)2 = x2 - 16 = (x)2 - (4)2
(a - b)2 ≠ a2 - b2
∴ (x-4)2 ≠ (x)2 - (4)2
∴ The given sentence is wrong.
Correct sentence is (x - 4)2 = x2 - 8x + 16.

(xii) (x + 7)2 = x2 +49
(x + 7)2 = x2 + 49 = (x)2 + (7)2
(a + b)2 ≠ a2 + b2
∴ (x+7)2 ≠ (x)2 - (7)2
∴ The given sentence is wrong.
Correct sentence is (x + 7)2 = x2 + 14x + 49.
(xiii) (3a + 4b)(a - b)= 3a2 - 4a2
3a(a - b) + 4b(a - b) = 3a2 - 42
3a2 - 3ab + 4ab - 4b2 = - a2
3a2 + ab - 4b2 ≠ a2
∴ The given sentence is wrong. Correct sentence is
(3a + 4b) (a - b) = 3a2 + ab - 4b2
(xiv) (x + 4) (x + 2) = x2 + 8
(x + 4) (x + 2) = x2 + 8
⇒ x2 + 6x + 8 = x2 + 8
⇒ 6x ≠ 0
Here 6x term is missing in R.H.S.
∴ The given sentence is wrong. Correct sentence is
(x + 4)(x + 2) = x2 + 6x + 8.
(xv) (x - 4) (x - 2) = x2- 8
(x - 4) (x - 2) = x2 - 8
⇒ x2 - 6x + 8 ≠ x2 - 8
∴ The given sentence is wrong. Correct sentence is
(x - 4) (x - 2) = x2 - 6x + 8