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

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex
14.1 are the part of NCERT Solutions for Class 8 Maths. Here you can find the
NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1**.**** **

**NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.1****NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.2****NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.3****NCERT Solutions for Class 8 Maths Chapter 14 Factorisation Ex 14.4**

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

Find the common factors of the given terms:**(i)**12x, 36

**(ii)**2y, 22xy

**(iii)**14pqr, 28p

^{2}q

^{2}

**(iv)**2x, 3x

^{2}, 4

**(v)**6abc, 24ab

^{2}, 12a

^{2}b

**(vi)**16x

^{3}, – 4x

^{2}, 32x

**(vii)**10pq, 20qr, 30rp

**(viii)**3x

^{2}y

^{3}, 10x

^{3}y

^{2}, 6x

^{2}y

^{2}z

**Solution:**

**(i)** The numerical coefficients in the given monomials are 12 and
36.

The highest common factor of 12 and 36 is 12.

But there is no common variable appearing in the given monomials 12x and 36.

Hence, the common factor of 12x and 36 is 12.

**(ii)** 2y, 22xy

= (2 × y) and (2 × 11 × x × y)

Common factors are 2 × y = 2y

Hence, the common factors of 2y and 22xy are 2y.

**(iii)** 14pq, 28p^{2}q^{2}

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

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

Hence, the common factors of 14pq and 28p^{2}q^{2} are 14pq.

**(iv)** 2x, 3x^{2}, 4

= (2 × x), (3 × x × x) and (2 × 2)

Common factor is 1.

Hence, the common factor of 2x, 3x^{2} and 4 is 1. [∵ 1 is a factor of every number]

**(v)** 6abc, 24ab^{2}, 12a^{2}b

= (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 factors of 6abc, 24ab^{2} and 12a^{2}b are
6ab.

**(vi)** 16x^{3}, -4x^{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 factors of 16x^{3}, -4x^{2} and 32x are 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 of 10pq, 20qr and 30rp is 10.

**(viii)** 3x^{2}y^{2}, 10x^{3}y^{2},
6x^{2}y^{2}z

= (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 = x^{2}y^{2}

Hence, the common factors of 3x^{2}y^{2}, 10x^{3}y^{2}
and 6x^{2}y^{2}z are x^{2}y^{2}.

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

Factorise the following
expressions.**(i)**7x – 42

**(ii)**6p – 12q

**(iii)**7a

^{2}+ 14a

**(iv)**-16z + 20z

^{3}

**(v)**20l

^{2}m + 30alm

**(vi)**5x

^{2}y – 15xy

^{2}

**(vii)**10a

^{2}– 15b

^{2}+ 20c

^{2}

**(viii)**-4a

^{2}+ 4ab – 4ca

**(ix)**x

^{2}yz + xy

^{2}z + xyz

^{2}

**(x)**ax

^{2}y + bxy

^{2}+ cxyz

**Solution:
(i)** 7x – 42 = 7(x – 6)

**(ii)**6p – 12q = 6(p – 2q)

**(iii)**7a

^{2}+ 14a = 7a(a + 2)

**(iv)**-16z + 20z

^{3}= 4z(-4 + 5z

^{2})

**(v)**20l

^{2}m + 30alm = 10lm(2l + 3a)

**(vi)**5x

^{2}y – 15xy

^{2}= 5xy(x – 3y)

**(vii)**10a

^{2}– 15b

^{2}+ 20c

^{2}= 5(2a

^{2}– 3b

^{2}+ 4c

^{2})

**(viii)**-4a

^{2}+ 4ab – 4ca = 4a(-a + b – c)

**(ix)**x

^{2}yz + xy

^{2}z + xyz

^{2}= xyz(x + y + z)

**(x)**ax

^{2}y + bxy

^{2}+ cxyz = xy(ax + by + cz)

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

Factorise:**(i)**x

^{2}+ 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)** x

^{2}+ xy + 8x + 8y

By grouping the terms, we get

(x

^{2}+ xy) + (8x + 8y)

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

= (x + y)(x + 8)

Hence, the required factorisation is (x + y)(x + 8).

**(ii)** 15xy – 6x + 5y – 2

By grouping the terms, we get

(15xy – 6x) + (5y – 2)

= 3x(5y – 2) + 1(5y – 2)

= (5y – 2)(3x + 1)

Hence, the required factorisation
is (5y – 2)(3x + 1).

**(iii)** ax + bx – ay – by

By grouping the terms, we get

(ax – ay) + (bx – by)

= a(x – y) + b(x – y)

= (x – y)(a + b)

Hence, the required factorisation is (x – y)(a + b).

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

By grouping the terms, we get

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

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

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

Hence, the required factorisation is (3q + 5) (5p + 3).

**(v) **z – 7 + 7xy – xyz

By grouping the terms, we get

(-xyz + 7xy) + (z – 7)

= -xy(z – 7) + 1(z – 7)

= (-xy + 1) (z – 7)

Hence, the required factorisation is (1 – xy) (z – 7).

**You can also like these:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**