Factorise 25abc2 – 15a2b2c

(i) 25abc² – 15a²b²c

(ii) 28p²q²r – 42pq²r²

Sol: 

(i) 25abc² – 15a²b²c

Take out common in both terms,

Then, 5abc (5c – 3ab)

Therefore, HCF of 25abc² and 15a²b²c is 5abc.

(ii) 28p²q²r – 42pq²r²

28p²q²r – 42pq²r²

Take out common in both terms,

Then, 14pq²r (2p – 3r)

Therefore, HCF of 28p²q²r and 42pq²r² is 14pq²r.

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