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**Factorisation Class 8 Extra Questions**

**NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Exercise 14.1**

**Ex 14.1 Class 8 Maths Question 1.**

Find the common factors of the given terms.

(i) 12x, 36

(ii) 2y, 22xy

(iii) 14pq, 28p2q2

(iv) 2x, 3×2, 4

(v) 6abc, 24ab2, 12a2b

(vi) 16×3, -4×2, 32x

(vii) 10pq, 20qr, 30rp

(viii) 3x2y3, 10x3y2, 6x2y2z**Solution**:

(i) 12x, 36

(2 × 2 × 3 × x) and (2 × 2 × 3 × 3)

Common factors are 2 × 2 × 3 = 12

Hence, the common factor = 12

(ii) 2y, 22xy

= (2 × y) and (2 × 11 × x × y)

Common factors are 2 × y = 2y

Hence, the common factor = 2y

(iii) 14pq, 28p2q2

= (2 × 7 × p × q) and (2 × 2 × 7 × p × p × q × q)

Common factors are 2 × 7 × p × q = 14pq

Hence, the common factor = 14pq

(iv) 2x, 3×2, 4

= (2 × x), (3 × x × x) and (2 × 2)

Common factor is 1

Hence, the common factor = 1 [∵ 1 is a factor of every number]

(v) 6abc, 24ab2, 12a2b

= (2 × 3 × a × b × c), (2 × 2 × 2 × 3 × a × b × b) and (2 × 2 × 3 × a × a × b)

Common factors are 2 × 3 × a × b = 6ab

Hence, the common factor = 6ab

(vi) 16×3, -4×2, 32x

= (2 × 2 × 2 × 2 × x × x × x), -(2 × 2 × x × x), (2 × 2 × 2 × 2 × 2 × x)

Common factors are 2 × 2 × x = 4x

Hence, the common factor = 4x

(vii) 10pq, 20qr, 30rp

= (2 × 5 × p × q), (2 × 2 × 5 × q × r), (2 × 3 × 5 × r × p)

Common factors are 2 × 5 = 10

Hence, the common factor = 10

(viii) 3x2y2, 10x3y2, 6x2y2z

= (3 × x × x × y × y), (2 × 5 × x × x × x × y × y), (2 × 3 × x × x × y × y × z)

Common factors are x × x × y × y = x2y2

Hence, the common factor = x2y2.

**Ex 14.1 Class 8 Maths Question 2.**

Factorise the following expressions.

(i) 7x – 42

(ii) 6p – 12q

(iii) 7a2 + 14a

(iv) -16z + 20z3

(v) 20l2m + 30alm

(vi) 5x2y – 15xy2

(vii) 10a2 – 15b2 + 20c2

(viii) -4a2 + 4ab – 4ca

(ix) x2yz + xy2z + xyz2

(x) ax2y + bxy2 + cxyz**Solution**:

(i) 7x – 42 = 7(x – 6)

(ii) 6p – 12q = 6(p – 2q)

(iii) 7a2 + 14a = 7a(a + 2)

(iv) -16z + 20z3 = 4z(-4 + 5z2)

(v) 20l2m + 30alm = 10lm(2l + 3a)

(vi) 5x2y – 15xy2 = 5xy(x – 3y)

(vii) 10a2 – 15b2 + 20c2 = 5(2a2 – 3b2 + 4c2)

(viii) -4a2 + 4ab – 4ca = 4a(-a + b – c)

(ix) x2yz + xy2z + xyz2 = xyz(x + y + z)

(x) ax2y + bxy2 + cxyz = xy(ax + by + cz)

**Ex 14.1 Class 8 Maths Question 3.**

Factorise:

(i) x2 + xy + 8x + 8y

(ii) 15xy – 6x + 5y – 2

(iii) ax + bx – ay – by

(iv) 15pq + 15 + 9q + 25p

(v) z – 7 + 7xy – xyz**Solution**:

(i) x2 + xy + 8x + 8y

Grouping the terms, we have

x2 + xy + 8x + 8y

= x(x + y) + 8(x + y)

= (x + y)(x + 8)

Hence, the required factors = (x + y)(x + 8)

(ii) 15xy – 6x + 5y – 2

Grouping the terms, we have

(15xy – 6x) + (5y – 2)

= 3x(5y – 2) + (5y – 2)

= (5y – 2)(3x + 1)

(iii) ax + bx – ay – by

Grouping the terms, we have

= (ax – ay) + (bx – by)

= a(x – y) + b(x – y)

= (x – y)(a + b)

Hence, the required factors = (x – y)(a + b)

(iv) 15pq + 15 + 9q + 25p

Grouping the terms, we have

= (15pq + 25p) + (9q + 15)

= 5p(3q + 5) + 3(3q + 5)

= (3q + 5) (5p + 3)

Hence, the required factors = (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz

Grouping the terms, we have

= (-xyz + 7xy) + (z – 7)

= -xy(z – 7) + 1 (z – 7)

= (-xy + 1) (z – 1)

Hence the required factor = -(1 – xy) (z – 7)

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