8th Standard, Mathematics, Chapter 12
Factorisation
Exercise 12.1
1 . Find the common factors of the given items.
(i ) 12x, 36
Answer: (i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
Common factors = 2, 2, 3.
2 × 2 × 3 = 12
(ii) 2y, 22xy
Answer: 2y = 2 × y
22xy = 2 × 11 × x × y
The common factors = 2, y.
2 × y = 2y
(iii) 14pq, 28p2q2
Answer: 14pq = 2 × 7 × p × q
28p2q2 = 2 × 2 × 7 × p × p × q × q
The common factors = 2, 7, p, q.
2 × 7 × p × q = 14pq
(iv) 2x, 3x2
Answer: 2x = 2 × x
3x2 = 3 × x × x
4 = 2 × 2
The common factor = 1.
(v) 6abc, 24ab2, 12a2b
Answer: 6abc = 2 × 3 × a × b × c
24ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
The common factors = 2, 3, a, b.
2 × 3 × a × b = 6ab
(vi) 16x3, -4x2, 32x
Answer: 16x3 = 2 × 2 × 2 × 2 × x × x × x
-4x2 = -1 × 2 × 2 × x × x
32x = 2 × 2 × 2 × 2 × 2 × x
The common factors = 2, 2, x.
2 × 2 × x = 4x
(vii) 10pq, 20qr, 30rp
Answer: 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
The common factors = 2, 5.
2 × 5 = 10
(viii) 3x2y3, ,10x3y2, ,6x2y2z
Answer: 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 = x, x, y, y.
Thus, x × x × y × y = x2 y2
2. Factorise the following expressions.
( i) 7x – 42
Answer: 7x – 42
7x = 7 × x
42 = 2 × 3 × 7
The common factor = 7.
7x – 42 = (7 × x) – (2 × 3 × 7 )
= 7(x – 6)
(ii) 6p – 12q
Answer: 6p -12q
6p = 2 × 3 × p
12q = 2 × 2 × 3 × q
The common factors = 2 and 3.
6p -12q = (2 × 3 × p) – (2 × 2 × 3 × q)
= 2 × 3(p – 2 × q)
= 6(p – 2q)
(iii) 7a2+14a
Answer: 7a2+14a
7a2= 7 × a × a
14a = 2 × 7 × a
The common factors = 7 and a .
7a2 + 14a = (7 × a × a) + (2 × 7 × a)
= 7 × a(a + 2)
= 7a (a + 2)
(iv) -16z + 20z3
Answer: -16z + 20z3
16z = -1 × 2 × 2 × 2 × 2 × z
20z3 = 2 × 2 × 5 × z × z × z
The common factors = 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 + 30alm
Answer: 20l2m + 30alm
20l2m= 2 × 2 × 5 × l × l × m
30alm = 2 × 3 × 5 × a × l × m
The common factors = 2, 5, l and m.
Therefore, 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 -15xy2
Answer: 5x2y -15xy2
5x2y = 5 × x × x × y
15xy2 = 3 × 5 × x × y × y
Common factors = 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 -15b2 + 20c2
Answer: 10a2 -15b2 + 20c2
10a2 = 2 × 5 × a × a
15b2 = 3 × 5 × b × b
20c2 = 2 × 2 × 5 × c × c
Common factor = 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 + 4ab – 4ca
Answer: -4a2 + 4ab – 4ca
4a2 = 2 × 2 × a × a
4ab = 2 × 2 × a × b
4ca = 2 × 2 × c × a
Common factors = 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 + xy2z + xyz2
Answer: x2yz + xy2z + xyz2
x2yz = x × x × y × z
xy2z= x × y × y × z
xyz2 = x × y × z × z
Common factors = 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 + bxy2 + cxyz
Answer: ax2y + bxy2 + cxyz
ax2y= a × x × x × y
bxy2 = b × x × y × y
cxyz= c × x × y × z
Common factors = 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)
Exercise 12.2
1. Factorise the following expressions.
(i) a2 + 8a +16
Answer: a2 + 8a +16
= (a)2 + 2 × a × 4 + (4)2
= (a + 4)2 [Using identity (x + y)2 = x2 + 2xy + y2, considering x = a and y = 4 ]
(ii) p2 – 10 p + 25
Answer: p2 – 10 p + 25
= (p)2 – 2 × p × 5 + (5)2
= (p – 5)2 [Using identity (a – b)2 = a2 – 2ab + b2, considering a = p and b = 5 ]
(iii) 25m2 + 30m + 9
Answer: 25m2 + 30m + 9
= (5m)2 + 2 × 5m × 3 + (3)2
= (5m + 3)2 [Using identity (a + b)2 = a2 + 2ab + b2, considering a = 5m and b = 3 ]
(iv) 49y2 + 84yz + 36z2
Answer: 49y2 + 84yz + 36z2
= (7y)2 + 2 × (7y) × (6z) + (6z)2
= (7y + 6z)2 [Using identity (a + b)2 = a2 + 2ab + b2, considering a = 7y and b = 6z]
(v) 4x2 – 8x+ 4
Answer: 4x2 – 8x+ 4
= (2x)2 – 2(2x)(2) + (2)2
= (2x – 2)2 [Using identity (a – b)2 = a2 – 2ab + b2, considering a = 2x and b = 2]
= [(2)(x -1)]2 = 4(x -1)2
(vi) 121b2 – 88bc+16c2
Answer: 121b2 – 88bc+16c2
= (11b)2 – 2(11b)(4c) + (4c)2
= (11b – 4c)2
(vii) (l + m)2 – 4lm
Answer: (l + m)2 – 4lm
= l2 + 2lm+ m2 – 4lm [Using identity (a + b)2 = a2 + 2ab + b2]
= l2 – 2lm + m2
= (l – m)2
(viii) a4 + 2a2b2 + b4
Answer: a4 + 2a2b2 + b4
= (a2)2 + 2 (a2)(b2) + (b2)2
= (a2 + b2)2 [Using identity (x + y)2 = x2 + 2xy + y2, considering x = a2 and y = b2]
2. Factorise.
(i) 4p2 – 9q2
Answer: 4p2 – 9q2
= (2p)2 – (3q)2
= (2p + 3q) (2p – 3q)
(ii) 63a2 – 112b2
Answer: 63a2 – 112b2
= 7(9a2 – 16b2 )
= 7 [(3a)2 – (4b)2]
= 7[(3a + 4b)(3a – 4b)] [Using identity x2 – y2 = (x – y)(x + y), considering x = 3a and y = 4b]
(iii) 49x2 – 36
Answer: 49x2 – 36
= (7x)2 – (6)2
= (7x – 6)(7x + 6)
(iv) 16x5 – 144x3
Answer: 16x5 – 144x3
= 16x3(x2 – 9)
= 16x3 [(x)2 – (3)2]
= 16x3 [(x – 3)(x + 3)] [Using identity a2 – b2 = (a – b)(a + b), considering a = x and b = 3]
(v) (l + m)2 – (l – m)2
Answer: (l + m)2 – (l – m)2
= [(l + m) – (l – m)][(l + m) + (l – m)]
= (l + m – l + m)(l + m + l – m)
= 2m × 2l
= 4ml
= 4lm
(vi) 9x2y2 – 16
Answer: 9x2y2 – 16
= (3xy)2 – (4)2
= (3xy – 4)(3xy + 4)
(vii) (x2 – 2xy + y2 ) – z2
Answer: (x2 – 2xy + y2 ) – z2
= (x – y)2 – (z)2
= (x – y – z)( x – y + z)
(viii) 25a2 – 4b2 + 28bc – 49c2
Answer: 25a2 – 4b2 + 28bc – 49c2
= 25a2 – (4b2 – 28bc + 49c2 )
= (5a)2 – [(2b)2 – 2 × 2b × 7c + (7c)2]
= (5a)2 – (2b – 7c)2 [Using identity (x – y)2 = x2 – 2xy + y2, considering x = 2b and y = 7c]
= [5a + (2b – 7c)][5a – (2b – 7c)]
= (5a + 2b – 7c)(5a – 2b + 7c)
3. Factorise the expressions.
(i) ax2 + bx
Answer: ax2 + bx
= a × x × x + b × x
= x (ax + b)
(ii) 7p2 + 21q2
Answer: 7p2 + 21q2
= 7 × p × p + 3 × 7 × q × q
= 7(p2 + 3q2)
(iii) 2x3 + 2xy2 + 2xz2
Answer: 2x3 + 2xy2 + 2xz2
= 2x( x2 + y2 + z2 )
(iv) am2 + bm2 + bn2 + an2
Answer: am2 + bm2 + bn2 + an2
= am2 + bm2 + an2 + bn2
= m2(a + b) + n2(a + b)
= (a + b)(m2 + n2 )
(v) (lm + l ) + m + 1
Answer: (lm + l ) + m + 1
= lm + m + l + 1
= m(l + 1) + 1(l + 1)
= (l + 1)(m + 1)
(vi) y (y + z) + 9(y + z)
Answer: y (y + z) + 9(y + z)
= (y + z) (y + 9)
(vii) 5y2 – 20 y – 8z + 2yz
Answer: 5y2 – 20 y – 8z + 2yz
= 5y2 – 20y + 2yz – 8z
= 5y(y – 4) + 2z(y – 4)
= (y – 4)(5y + 2z)
(viii) 10ab + 4a + 5b + 2
Answer: 10ab + 4a + 5b + 2
= 10ab + 5b + 4a + 2
= 5b(2a +1) + 2(2a +1)
= (2a +1)(5b + 2)
(ix) 6xy – 4 y + 6 – 9x
Answer: 6xy – 4 y + 6 – 9x
= 6xy – 9x – 4 y + 6
= 3x(2y – 3) – 2(2 y – 3)
= (2y – 3)(3x – 2)
4. Factorise.
(i) a4 – b4
Answer: a4 – b4
= (a2)2 – (b2)2
= (a2 – b2)(a2 + b2) [Since, a2 – b2 = (a – b)(a + b)]
= (a – b)(a + b)(a2 + b2 ) [Since, a2 – b2 = (a – b)(a + b)]
(ii) p4 – 81
Answer: p4 – 81
= (p2)2 – (9)2
= (p2 – 9)(p2 + 9) [Since, a2 – b2 = (a – b)(a + b)]
= [(p)2 – (3)2](p2 + 9)
= (p – 3)(p + 3) (p2 + 9) [Since, a2 – b2 = (a – b)(a + b)]
(iii) x4 – (y + z)4
Answer: x4 – (y + z)4
= (x2)2 – [(y + z)2]2
= [x2 – (y + z)2][x2 + (y + z)2] [Since, a2 – b2 = (a – b)(a + b)]
= [x – ( y + z)][x + ( y + z)][x2 + (y + z)2] [Since, a2 – b2 = (a – b)(a + b)]
= (x – y – z)(x + y + z)[x2 + (y + z)2]
(iv) x4 – (x – z)4
Answer: x4 – (x – z)4
= (x2)2 – [( x – z)2]2
= [x2 – (x – z)2 ][ x2 + (x – z)2] [Since, a2 – b2 = (a – b)(a + b)]
= [x – (x – z)][x + (x – z )][x2 + ( x – z )2] [Since, a2 – b2 = (a – b)(a + b)]
= z(2x – z )[x2 + x2 – 2xz + z2] [Since, (a- b)2 = a2 – 2ab+ b2]
= z(2x – z )(2x2 – 2xz + z2)
(v) a4 – 2a2b2 + b4
Answer: a4 – 2a2b2 + b4
= (a2)2 – 2 (a2)(b2) + (b2)2
= (a2 – b2)2 [Since, (a- b)2 = a2 – 2ab+ b2]
= [(a – b)(a + b)]2 [Since, a2 – b2 = (a – b)(a + b)]
= (a – b)2(a + b)2
5. Factorise the following expressions.
( i) p2 + 6 p + 8
Answer: p2 + 6 p + 8
8 = 4 × 2 and 4 + 2 = 6
p2 + 6 p + 8 = p2 + 2p + 4p + 8
= p(p + 2) + 4(p + 2)
= (p + 2) (p + 4)
(ii) q2 – 10q + 21
Answer: q2 – 10q + 21
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 + 6 p -16
Answer: p2 + 6 p -16
It can be observed that, -16 = (-2) × 8 and 8 + (-2) = 6
p2 + 6 p -16 = p2 + 8p – 2p -16
= p(p + 8) – 2(p + 8)
= (p + 8)(p – 2)
Exercise 12.3
1 . Carry out the following divisions.
(i) 28x4 ÷ 56x
Answer: 28x4 = 2 × 2 × 7 × x × x × x × x and
56x = 2 × 2 × 2 × 7 × x
28x4 ÷ 56x = (2 × 2 × 7 × x × x × x × x) / (2 × 2 × 2 × 7 × x)
= x3/2
(ii) – 36y3 ÷ 9y2
Answer: -36y3 = – 2 × 2 × 3 × 3 × y × y × y and
9y2 = 3 × 3 × y × y
-36y3 ÷ 9y2 = (-2 × 2 × 3 × 3 × y × y × y) / (3 × 3 × y × y)
= -4y
(iii) 66 pq2r3 ÷ 11qr2
Answer: 66pq2r3 = 2 × 3 × 11 × p × q × q × r × r × r and
11qr2 = 11 × q × r × r
66 pq2r3 ÷ 11qr2 = (2 × 3 × 11 × p × q × q × r × r × r) / (11 × q × r × r)
= 6pqr
(iv) 34x3y3z3 ÷ 51xy2z3
Answer: 34x3y3z3 = 2 × 17 × x × x × x × y × y × y × z × z × z and
51xy2z3 = 3 × 17 × x × y × y × z × z × z
34x3y3z3 ÷ 51xy2z3 = (2 × 17 × x × x × x × y × y × y × z × z × z) / (3 × 17 × x × y × y × z × z × z)
= 2x2y / 3
(v) 12a8b8 ÷ (-6a6b4)
Answer: 12a8b8 = 2 × 2 × 3 × a8 × b8 and
-6a6b4 = -2 × 3 × a6 × b4
12a8b8 ÷ (-6a6b4) = (2 × 2 × 3 × a8 × b8) / (-2 × 3 × a6 × b4)
= -2a2b4
2. Divide the given polynomial by the given monomial.
(i) (5x2 – 6x) ÷ 3x
Answer: (5x2 – 6x) = x(5x – 6)
(5x2 – 6x) ÷ 3x = x(5x – 6) / 3x
= (5x – 6) / 3
(ii) (3y8 – 4y6 + 5y4) ÷ y4
Answer: (3y8 – 4y6 + 5y4) = y4(3y4 – 4y2 + 5)
(3y8 – 4y6 + 5y4) ÷ y4 = y4(3y4 – 4y2 + 5) / y4
= 3y4 – 4y2 + 5
(iii) 8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2
Answer: 8(x3y2z2 + x2y3z2 + x2y2z3) = 8x2y2z2(x + y + z)
8(x3y2z2 + x2y3z2 + x2y2z3) ÷ 4x2y2z2 = 8x2y2z2(x + y + z) / 4x2y2z2
= 2(x + y + z)
(iv) (x3 + 2x2 + 3x) ÷ 2x
Answer: (x3 + 2x2 + 3x) = x(x2 + 2x + 3)
(x3 + 2x2 + 3x) ÷ 2x = x(x2 + 2x + 3) / 2x
= (x2 + 2x + 3) / 2
(v) (p3q6 – p6q3) ÷ p3q3
Answer; (p3q6 – p6q3) = p3q3 (q3 – p3)
(p3q6 – p6q3) ÷ p3q3= p3q3 (q3 – p3) / p3q3
= q3 – p3
3. Work out the following divisions.
(i) (10x – 25) ÷ 5
Answer: (10x – 25) = 5× 2 × x – 5 × 5
= 5(2x – 5)
(10x – 25) ÷ 5 = 5(2x – 5) / 5
= 2x – 5
(ii) (10x – 25) ÷ (2x – 5)
Answer: (10x – 25) = 5 × 2 × x – 5× 5
= 5(2x – 5)
(10x – 25) ÷ (2x – 5) = 5(2x – 5) / (2x – 5)
= 5
(iii) 10y (6y + 21) ÷ 5(2y + 7)
Answer: 10y(6y + 21) = 5 × 2 × y ×(2 × 3 × y + 3 × 7)
= 5 × 2 × y × 3(2 × y + 7)
= 30y(2y + 7)
10y(6y + 21) ÷ 5(2y + 7) = 30y(2y + 7) / 5(2y + 7)
= 6y
(iv) 9x2y2 (3z – 24) ÷ 27xy(z – 8)
Answer: 9x2y2 (3z – 24) = 3 × 3 × x × x × y × y ×(3 × z – 2 × 2 × 2 × 3)
= 3 × 3 × x × x × y × y × 3 (z – 2 × 2 × 2)
= 27x2y2 (z – 8)
9x2y2 (3z – 24) ÷ 27xy(z – 8) = 27x2y2(z – 8) / 27xy(z – 8)
= xy
(v) 96abc (3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6)
Answer: 96abc (3a – 12) (5b – 30) = 96abc ×(3 × a – 2 × 2 × 3) × (5 × b – 5 × 2 × 3)
= 96abc × 3(a – 2 × 2) × 5(b – 2 × 3)
= 1440abc (a – 4) (b – 6)
96abc (3a – 12)(5b – 30) ÷ 144(a – 4)(b – 6) = 1440abc(a – 4)(b – 6) / 144(a – 4)(b – 6)
= 10abc
4. Divide as directed.
(i) 5(2x +1) (3x + 5) ÷ (2x +1)
Answer: 5(2x +1) (3x + 5) / (2x +1)
= 5(3x + 5)
(ii) 26xy(x + 5) (y – 4) ÷ 13x(y – 4)
Answer: 2 × 13 × xy(x + 5) (y – 4) / 13x(y – 4)
= 2y(x + 5)
(iii) 52pqr (p + q) (q + r)(r + p) ÷ 104pq(q + r)(r + p)
Answer: 2 × 2 × 13 × p × q × r × (p + q) × (q + r) × (r + p) / 2 × 2 × 2 × 13 × p × q × (q + r) × (r + p)
= r (p + q) / 2
(iv) 20 (y + 4) (y2 + 5y + 3) ÷ 5(y + 4)
Answer: 2 × 2 × 5 × (y + 4) × (y2 + 5 y + 3) / 5 × (y + 4)
= 4(y2 + 5 y + 3)
(v) x (x +1)(x + 2)(x + 3) ÷ x(x +1)
= (x + 2) (x + 3)
5. Factorise the expressions and divide them as directed.
(i) (y2 + 7y + 10) ÷ (y + 5)
Answer: (y2 + 7y + 10) = y2 + 2 y + 5 y +10 = y(y + 2) + 5(y + 2)
= (y + 2)(y + 5)
(y2 + 7y + 10) ÷ (y + 5) = (y + 2)(y + 5) / (y + 5)
= y + 2
(ii) (m2 -14m – 32) ÷ (m + 2)
Answer: m2 + 2m -16m – 32 = m(m + 2) -16(m + 2)
= (m + 2)(m – 16)
(m2 -14m – 32) ÷ (m + 2) = (m + 2)(m – 16) / (m + 2)
= m – 16
(iii) (5p2 – 25p + 20) ÷ (p -1)
Answer: 5(p2 – 5p + 4) = 5(p2 – p – 4p + 4)
= 5[p (p – 1) – 4(p – 1)]
= 5(p – 1) (p – 4)
(5p2 – 25p + 20) ÷ (p – 1) = 5(p – 1)(p – 4) / (p -1)
= 5(p – 4)
(iv) 4yz (z2 + 6z -16) ÷ 2y (z + 8)
Answer: 4 yz (z2 – 2z + 8z -16) = 4 yz [z(z – 2) + 8(z – 2)]
= 4 yz (z – 2)(z + 8)
4yz (z2 + 6z -16) ÷ 2y (z + 8) = 4 yz(z – 2)(z + 8) / 2y(z + 8)
= 2z (z – 2)
(v) 5pq (p2 – q2) ÷ 2p (p + q)
Answer: 5pq (p – q) (p + q) [Using identity a2 – b2 = (a + b)(a – b)]
Thus, 5pq (p2 – q2) ÷ 2p (p + q) = 5pq(p – q)(p + q) / 2p(p + q)
= 5q (p – q) / 2
(vi) 12xy (9x2 -16y2) ÷ 4xy (3x + 4y)
Answer: 12xy [(3x)2 – (4y)2] = 12xy(3x – 4y)(3x + 4y) [Using identity a2 – b2 = (a + b)(a – b)]
= 2 × 2 × 3 × x × y × (3x – 4y) × (3x + 4y)
12xy (9x2 -16y2) ÷ 4xy (3x + 4y) = 2 × 2× 3 × x × y × (3x – 4y) × (3x + 4y) / 4xy (3x + 4y)
= 3(3x – 4y)
(vii) 39y3 (50y2 – 98) ÷ 26y2 (5y + 7)
Answer: 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) [Using identity a2 – b2 = (a + b) (a – b)]
26y2 (5y + 7) = 2 × 13 × y × y × (5y + 7)
39y3 (50y2 – 98) ÷ 26y2 (5y + 7) = 3 × 13 × 2 × y × y × y (5y – 7) (5y + 7) / 2 × 13 × y × y × (5y + 7)
= 3y (5y – 7)
