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Question -

Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy

(iii) 14pq, 28p2q2
(iv) 2x, 3x
2, 4
(v) 6abc, 24ab
2, 12a2b
(vi) 16x
3, -4x2, 32x
(vii) 10pq, 20qr, 30rp
(viii) 3x
2y3, 10x3y2, 6x2y2z



Answer -

(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 afactor 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 = x
2y2
Hence, the common factor = x
2y2.

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