Question

1. Find the common factors of the given terms
(1) 12x, 36why

(1) 2y, 22x (i) 14 pu. 28p-9
(iv) 2x, 3x-, 4 (v) 6 abc, 24ab2, 12 ab
(vi) 16 x", - 4x2, 32x
(vii) 10 pq, 20qr, 30rp
(viii) 3x2 y3, 10x y2,6 x2 yzz​

Answer 1

: 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, 3x2, 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) 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 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.