# HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 Factorization Ex 14.1 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 14 Factorization Exercise 14.1

Question 1.
Find the common factors of the given terms:
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28 p2q2
(iv) 2x, 3x2, 4
(v) 6abc, 24 ab2, 12a2b
(vi) 16x3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x2y3, 10x3y2, 6x2y2z
Solution:
(i) 12x = 2 × 2 × 3 × x
36 = 2 × 2 × 3 × 3
∴ Highest common factor, HCF = 2 × 2 × 3 = 12

(ii) 2y = 2 × y
22xy = 2 × 11 × x × y
∴ HCF = 2 × y = 2y

(iii) 14pq = 2 × 7 × p x q
28 p2q2 = 2 × 2 × 7 × p × p × q × q
∴ HCF = 2 × 7 × p × q = 14 pq

(iv) 2x = 2 × x
3x2 = 3 × x × x
4 = 2 × 2
∴ HCF = 1

(v) 6abc = 2 × 3 × a × b × c
24 ab2 = 2 × 2 × 2 × 3 × a × b × b
12a2b = 2 × 2 × 3 × a × a × b
∴ HCF = 2 × 3 × a × b
= 6ab

(vi) 16x3 = 2 × 2 × 2 × 2 × x × x × x
-4x2 = (-2) × 2 × x
32x = 2 × 2 × 2 × 2 × 2 × x
∴ HCF = 2 × 2 × x
= 4x

(vii) 10pq = 2 × 5 × p × q
20qr = 2 × 2 × 5 × q × r
30rp = 2 × 3 × 5 × r × p
∴ HCF = 2 × 5
= 10

(viii) 3x2y3 = 3 × x × x × y × y × y
10 x3 y22 = 2 × 5 × x × x × x y × y
6x2y2z = 2 × 3 × x × x × y × y × z
∴ HCF = x × x × y × y
= x2y2

Question 2.
Factorise the following ex-pressions
(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) 7a – 42 = 7(x – 6)
(ii) 6p – 12q = 6 (p – 2q)
(iii) 7a2 + 14a = 7a (a + 2)
(iv) -16z + 20z3 = 42 (-4z + 5z2)
= 4z (5z2 – 4z)
(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)

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

(ii) 15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= x(a + b) – y (a + b)
= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p
= 15pq + 9q + 25p + 15
= 3q + (5p + 3) + 5 (5p + 3)
= (5p + 3) (3q + 5)

(v) z – 7 + 7xy – xyz
= z – 7 – xyz + 7xy
= 1 (z – 7) – xy (z – 7)
= (z – 7)(1 – xy)