1.

Factorise (ii) 14(a – 3b)3 – 21p(a – 3b)​

Answer»

Answer:

14(a−3b)

3

−21p(a−3b)

TAKE out common in all terms,

Then,7(a−3b)[2(a−3b)

2

−3p]

THEREFORE, HCF of 14(a−3b)

3

and 21p(a−3b) is 7(a−3b).


Discussion

No Comment Found

Related InterviewSolutions