# 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) 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