How do you factor completely $3p^3q-9pq^2+36pq$ ?

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Answer 1 · 2015-11-27T00:10:09

$3p3q−9pq2+36pq = 3pq(p2 - 3q + 12)$

What is present in every part of this equation? what is the lowest common factor of these?

Well we have a $p$ present in all of them, and we also have a $q$ in all of them as well.

Now is there another greatest common factor among $3$ , $9$ , and $36$ ? Yes! It is $3$ .

So we can divide every term in the equation by $3pq$

Take out $3pq$ by dividing every term by $3pq$ :
$3pq((3p3q)/(3pq)−(9pq2)/(3pq)+(36pq)/(3pq))$

So our final answer is $3pq(p2 - 3q + 12)$

How do you factor completely 3p^3q-9pq^2+36pq3p^3q-9pq^2+36pq?