Find the common factors of the given terms
(i) 12x, 36   (ii)2y, 22xy   (iii)14pq, 28p²q²   (iv)2x, 3x², 4
(v) 6abc, 24ab², 12a²b   (vi) 16x³, - 4x², 32x
(vii)10 pq, 20qr, 30rp   (viii)3x² y³, 10x³y², 6x²y²z

Solution:

We will first find out factors of each term then check which factors are common in each term.

(i) 12x = 2 × 2 × 3 × x

36 = 2 × 2 × 3 × 3

The common factors are 2, 2, 3.

Thus, 2 × 2 × 3 = 12

(ii) 2y = 2 × y

22xy = 2 × 11 × x × y

The common factors are 2, y.

Thus, 2 × y = 2y

(iii) 14pq = 2 × 7 × p × q

28p²q² = 2 × 2 × 7 × p × p × q × q

The common factors are 2, 7, p, q.

Thus, 2 × 7 × p × q = 14pq

(iv) 2x = 2 × x

3x² = 3 × x × x

4 = 2 × 2

The common factor is 1.

(v) 6abc = 2 × 3 × a × b × c

24ab² = 2 × 2 × 2 × 3 × a × b × b

12a²b = 2 × 2 × 3 × a × a × b

The common factors are 2, 3, a, b.

Thus, 2 × 3 × a × b = 6ab

(vi) 16x³ = 2 × 2 × 2 × 2 × x × x × x

-4x² = -1 × 2 × 2 × x × x

32x = 2 × 2 × 2 × 2 × 2 × x

The common factors are 2, 2, x.

Thus, 2 × 2 × x = 4x

(vii) 10pq = 2 × 5 × p × q

20qr = 2 × 2 × 5 × q × r

30rp = 2 × 3 × 5 × r × p

The common factors are 2, 5.

Thus, 2 × 5 = 10

(viii) 3x²y³ = 3 × x × x × y × y × y

10x³y² = 2 × 5 × x × x × x × y × y

6x²y²z = 2 × 3 × x × x × y × y × z

The common factors are x, x, y, y.

Thus, x × x × y × y = x² y²

☛ Check: NCERT Solutions for Class 8 Maths Chapter 14

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Video Solution:

Find the common factors of the terms (i) 12x, 36 (ii)2y, 22xy (iii)14pq, 28p²q² (iv)2x, 3x², 4 (v) 6abc, 24ab², 12a²b (vi) 16x³, - 4x², 32x (vii)10 pq, 20qr, 30rp (viii)3x² y³, 10x³y², 6x²y²z

Class 8 Maths NCERT Solutions Chapter 14 Exercise 14.1 Question 1

Summary:

The common factors of the terms (i) 12x, 36 (ii)2y, 22xy (iii)14pq, 28p²q² (iv)2x, 3x², 4 (v) 6abc, 24ab², 12a²b (vi) 16x³, - 4x², 32x (vii)10 pq, 20qr, 30rp (viii)3x² y³, 10x³y², 6x²y²z are as follows (i) 2, 2, 3 = 12 (ii) 2, y = 2y (iii) 2, 7, p, q = 14pq (iv) 1 (v) 2, 3, a, b = 6ab (vi) 2, 2, x = 4x (vii) 2, 5 = 10 (viii) x, x, y, y = x² y²

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