Factorise $7p^{2}q + 14pq^{2} - 21pq$.
Factorise $7p^{2}q + 14pq^{2} - 21pq$.
- $7pq(p + 2q - 3)$
- $7pq(p + 2q + 3)$
- $7(p^{2}q + 2pq^{2} - 3pq)$
- $pq(7p + 14q - 21)$
Solution
Common factor is $7pq$. Dividing each term: $\frac{7p^{2}q}{7pq}=p$, $\frac{14pq^{2}}{7pq}=2q$, $\frac{-21pq}{7pq}=-3$. So $7pq(p+2q-3)$.
Asked in: IMO
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