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 14.1

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.

### Ex 14.1 Class 8 Maths Question 1.

Find the common factors of the given terms:
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pqr, 28p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24ab2, 12a2b
(vi) 16x3, – 4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z

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, 28p2q2
= (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 28p2q2 are 14pq.

(iv) 2x, 3x2, 4
= (2 × x), (3 × x × x) and (2 × 2)
Common factor is 1.
Hence, the common factor of 2x, 3x2 and 4 is 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 factors of 6abc, 24ab2 and 12a2b are 6ab.

(vi) 16x3, -4x2, 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 16x3, -4x2 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) 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 factors of 3x2y2, 10x3y2 and 6x2y2z are 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
By grouping the terms, we get
(x2 + 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).

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