# Factorisation Class 8 Extra Questions Maths Chapter 14

Extra Questions for Class 8 Maths Chapter 14 Factorisation

### Factorization Class 8 Extra Questions Very Short Answer Type

Question 1.
Find the common factors of the following terms.
(a) 25x2y, 30xy2
(b) 63m3n, 54mn4
Solution:
(a) 25x2y, 30xy2
25x2y = 5 × 5 × x × x × y
30xy2 = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy

(b) 63m3n, 54mn4
63m3n = 3 × 3 × 7 × m × m × m × n
54mn4 = 2 × 3 × 3 × 3 × m × n × n × n × n
Common factors are 3 × 3 × m × n = 9mn

Question 2.
Factorise the following expressions.
(a) 54m3n + 81m4n2
(b) 15x2y3z + 25x3y2z + 35x2y2z2
Solution:
(a) 54m3n + 81m4n2
= 2 × 3 × 3 × 3 × m × m × m × n + 3 × 3 × 3 × 3 × m × m × m × m × n × n
= 3 × 3 × 3 × m × m × m × n × (2 + 3 mn)
= 27m3n (2 + 3mn)

(b) 15x2y3z + 25 x3y2z + 35x2y2z2 = 5x2y2z ( 3y + 5x + 7)

Question 3.
Factorise the following polynomials.
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
Solution:
(a) 6p(p – 3) + 1 (p – 3) = (p – 3) (6p + 1)
(b) 14(3y – 5z)3 + 7(3y – 5z)2
= 7(3y – 5z)2 [2(3y – 5z) +1]
= 7(3y – 5z)2 (6y – 10z + 1)

Question 4.
Factorise the following:
(a) p2q – pr2 – pq + r2
(b) x2 + yz + xy + xz
Solution:
(a) p2q – pr2 – pq + r2
= (p2q – pq) + (-pr2 + r2)
= pq(p – 1) – r2(p – 1)
= (p – 1) (pq – r2)

(b) x2 + yz + xy + xz
= x2 + xy +xz + yz
= x(x + y) + z(x + y)
= (x + y) (x + z)

Question 5.
Factorise the following polynomials.
(a) xy(z2 + 1) + z(x2 + y2)
(b) 2axy2 + 10x + 3ay2 + 15
Solution:
(a) xy(z2 + 1) + z(x2 + y2)
= xyz2 + xy + 2x2 + zy2
= (xyz2 + zx2) + (xy + zy2)
= zx(yz + x) + y(x + yz)
= zx(x + yz) + y(x + yz)
= (x + yz) (zx + y)

(b) 2axy2 + 10x + 3ay2 + 15
= (2axy2 + 3ay2) + (10x + 15)
= ay2(2x + 3) +5(2x + 3)
= (2x + 3) (ay2 + 5)

Question 6.
Factorise the following expressions.
(а) x2 + 4x + 8y + 4xy + 4y2
(b) 4p2 + 2q2 + p2q2 + 8
Solution:
(a) x2 + 4x + 8y + 4xy + 4y2
= (x2 + 4xy + 4y2) + (4x + 8y)
= (x + 2y)2 + 4(x + 2y)
= (x + 2y)(x + 2y + 4)

(b) 4p2 + 2q2 + p2q2 + 8
= (4p2 + 8) + (p2q2 + 2q2)
= 4(p2 + 2) + q2(p2 + 2)
= (p2 + 2)(4 + q2)

Question 7.
Factorise:
(a) a2 + 14a + 48
(b) m2 – 10m – 56
Solution:
(a) a2 + 14a + 48
= a2 + 6a + 8a + 48
[6 + 8 = 14 ; 6 × 8 = 48]
= a(a + 6) + 8(a + 6)
= (a + 6) (a + 8)

(b) m2 – 10m – 56
= m2 – 14m + 4m – 56
[14 – 4 = 10; 4 × 4 = 56]
= m(m – 14) + 6(m – 14)
= (m – 14) (m + 6)

Question 8.
Factorise:
(a) x4 – (x – y)4
(b) 4x2 + 9 – 12x – a2 – b2 + 2ab
Solution:
(a) x4 – (x – y)4
= (x2)2 – [(x – y)2]2
= [x2 – (x – y)2] [x2 + (x – y)2]
= [x + (x – y] [x – (x – y)] [x2 + x2 – 2xy + y2]
= (x + x – y) (x – x + y)[2x2 – 2xy + y2]
= (2x – y) y(2x2 – 2xy + y2)
= y(2x – y) (2x2 – 2xy + y2)

(b) 4x2 + 9 – 12x – a2 – b2 + 2ab
= (4x2 – 12x + 9) – (a2 + b2 – 2ab)
= (2x – 3)2 – (a – b)2
= [(2x – 3) + (a – b)] [(2x – 3) – (a – b)]
= (2x – 3 + a – b)(2x – 3 – a + b)

### Factorisation Class 8 Extra Questions Short Answer Type

Question 9.
Factorise the following polynomials.
(a) 16x4 – 81
(b) (a – b)2 + 4ab
Solution:
(a) 16x4 – 81
= (4x2)2 – (9)2
= (4x2 + 9)(4x2 – 9)
= (4x2 + 9)[(2x)2 – (3)2]
= (4x2 + 9)(2x + 3) (2x – 3)

(b) (a – b)2 + 4ab
= a2 – 2ab + b2 + 4ab
= a2 + 2ab + b2
= (a + b)2

Question 10.
Factorise:
(а) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
Solution:
(a) 14m5n4p2 – 42m7n3p7 – 70m6n4p3
= 14m5n3p2(n – 3m2p5 – 5mnp)

(b) 2a2(b2 – c2) + b2(2c2 – 2a2) + 2c2(a2 – b2)
= 2a2(b2 – c2) + 2b2(c2 – a2) + 2c2(a2 – b2)
= 2[a2(b2 – c2) + b2(c2 – a2) + c2(a2 – b2)] = 2 × 0
= 0

Question 11.
Factorise:
(a) (x + y)2 – 4xy – 9z2
(b) 25x2 – 4y2 + 28yz – 49z2
Solution:
(a) (x + y)2 – 4xy – 9z2
= x2 + 2xy + y2 – 4xy – 9z2
= (x2 – 2xy + y2) – 9z2
= (x – y)2 – (3z)2
= (x – y + 3z) (x – y – 3z)

(b) 25x2 – 4y2 + 28yz – 49z2
= 25x2 – (4y2 – 28yz + 49z2)
= (5x)2 – (2y – 7)2
= (5x + 2y – 7) [5x – (2y – 7)]
= (5x + 2y – 7) (5x – 2y + 7)

Question 12.
Evaluate the following divisions:
(a) (3b – 6a) ÷ (30a – 15b)
(b) (4x2 – 100) ÷ 6(x + 5)
Solution: Question 13.
Simplify the following expressions: Solution: Question 14.
Factorise the given expressions and divide that as indicated.
(a) 39n3(50n2 – 98 ) ÷ 26n2(5n – 7)
(b) 44(p4 – 5p3 – 24p2) ÷ 11p(p – 8)
Solution:  Question 15.
If one of the factors of (5x2 + 70x – 160) is (x – 2). Find the other factor.
Solution:
Let the other factor be m.
(x – 2) × m = 5x2 + 70x – 160             +