Factorisation Class 8 Extra Questions Maths Chapter 14

Factorisation Class 8 Extra Questions Maths Chapter 14

Extra Questions for Class 8 Maths Chapter 14 Factorisation

Factorisation Class 8 Extra Questions Very Short Answer Type

Question 1.
Find the common factors of the following terms.
(a) 25x ² y, 30xy ²
(b) 63m ³ n, 54mn ⁴
Solution:
(a) 25x ² y, 30xy ²
25x ² y = 5 × 5 × x × x × y
30xy ² = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy

(b) 63m ³ n, 54mn ⁴
63m ³ n = 3 × 3 × 7 × m × m × m × n
54mn ⁴ = 2 × 3 × 3 × 3 × m × n × n × n × n
Common factors are 3 × 3 × m × n = 9mn

Question 2.
Factorise the following expressions.
(a) 54m ³ n + 81m ⁴ n ²
(b) 15x ² y ³ z + 25x ³ y ² z + 35x ² y ² z ²
Solution:
(a) 54m ³ n + 81m ⁴ n ²
= 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)
= 27m ³ n (2 + 3mn)

(b) 15x ² y ³ z + 25 x ³ y ² z + 35x ² y ² z ² = 5x ² y ² z ( 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) p ² q – pr ² – pq + r ²
(b) x ² + yz + xy + xz
Solution:
(a) p ² q – pr ² – pq + r ²
= (p ² q – pq) + (-pr ² + r2)
= pq(p – 1) – r ² (p – 1)
= (p – 1) (pq – r ² )

(b) x ² + yz + xy + xz
= x ² + xy +xz + yz
= x(x + y) + z(x + y)
= (x + y) (x + z)

Question 5.
Factorise the following polynomials.
(a) xy(z ² + 1) + z(x ² + y ² )
(b) 2axy ² + 10x + 3ay ² + 15
Solution:
(a) xy(z ² + 1) + z(x ² + y ² )
= xyz ² + xy + 2x ² + zy ²
= (xyz ² + zx ² ) + (xy + zy ² )
= zx(yz + x) + y(x + yz)
= zx(x + yz) + y(x + yz)
= (x + yz) (zx + y)

(b) 2axy ² + 10x + 3ay ² + 15
= (2axy ² + 3ay ² ) + (10x + 15)
= ay ² (2x + 3) +5(2x + 3)
= (2x + 3) (ay ² + 5)

Question 6.
Factorise the following expressions.
(а) x ² + 4x + 8y + 4xy + 4y ²
(b) 4p ² + 2q ² + p ² q ² + 8
Solution:
(a) x ² + 4x + 8y + 4xy + 4y ²
= (x ² + 4xy + 4y ² ) + (4x + 8y)
= (x + 2y) ² + 4(x + 2y)
= (x + 2y)(x + 2y + 4)

(b) 4p ² + 2q ² + p ² q ² + 8
= (4p ² + 8) + (p ² q ² + 2q ² )
= 4(p ² + 2) + q ² (p ² + 2)
= (p ² + 2)(4 + q ² )

Question 7.
Factorise:
(a) a ² + 14a + 48
(b) m ² – 10m – 56
Solution:
(a) a ² + 14a + 48
= a ² + 6a + 8a + 48
[6 + 8 = 14 ; 6 × 8 = 48]
= a(a + 6) + 8(a + 6)
= (a + 6) (a + 8)

(b) m ² – 10m – 56
= m ² – 14m + 4m – 56
[14 – 4 = 10; 4 × 4 = 56]
= m(m – 14) + 6(m – 14)
= (m – 14) (m + 6)

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

(b) 4x ² + 9 – 12x – a ² – b ² + 2ab
= (4x ² – 12x + 9) – (a ² + b ² – 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) 16x ⁴ – 81
(b) (a – b) ² + 4ab
Solution:
(a) 16x ⁴ – 81
= (4x ² ) ² – (9)2
= (4x ² + 9)(4x ² – 9)
= (4x ² + 9)[(2x) ² – (3) ² ]
= (4x ² + 9)(2x + 3) (2x – 3)

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

Question 10.
Factorise:
(а) 14m ⁵ n ⁴ p ² – 42m ⁷ n ³ p ⁷ – 70m ⁶ n ⁴ p ³
(b) 2a ² (b ² – c ² ) + b ² (2c ² – 2a ² ) + 2c ² (a ² – b ² )
Solution:
(a) 14m ⁵ n ⁴ p ² – 42m ⁷ n ³ p ⁷ – 70m ⁶ n ⁴ p ³
= 14m ⁵ n ³ p ² (n – 3m ² p ⁵ – 5mnp)

(b) 2a ² (b ² – c ² ) + b ² (2c ² – 2a ² ) + 2c ² (a ² – b ² )
= 2a ² (b ² – c ² ) + 2b ² (c ² – a ² ) + 2c ² (a ² – b ² )
= 2[a ² (b ² – c ² ) + b ² (c ² – a ² ) + c ² (a ² – b ² )]

= 2 × 0
= 0

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

(b) 25x ² – 4y ² + 28yz – 49z ²
= 25x ² – (4y ² – 28yz + 49z ² )
= (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) (4x ² – 100) ÷ 6(x + 5)
Solution:

Question 13.
Simplify the following expressions:

Solution:

Question 14.
Factorise the given expressions and divide that as indicated.
(a) 39n ³ (50n ² – 98 ) ÷ 26n ² (5n – 7)
(b) 44(p ⁴ – 5p ³ – 24p ² ) ÷ 11p(p – 8)
Solution:

Question 15.
If one of the factors of (5x ² + 70x – 160) is (x – 2). Find the other factor.
Solution:
Let the other factor be m.
(x – 2) × m = 5x ² + 70x – 160

Extra Questions for Class 8 Maths

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