If anyone could help I'll give them a medal! It will be posted below.

Question

anonymous · Answer

$$\frac{ m^2+mn-2n^2 }{ m^3+m^2n }\times \frac{ m^2+3mn+2n^2 }{ m }$$

larseighner · Answer

It would probably be useful to factor the numerators before doing anything else.

anonymous · Answer

im having a hard time factoring it all out.

anonymous · Answer

this is what i got. $$\frac{ mn(m-2n) }{ m^2(m+n)}\times \frac{ (m+n)(m+2n) }{ m }$$

larseighner · Answer

Well, just take one of them to begin with.

$$\large m^2 + mn -2n^2$$

The end terms are squared so the form of the factors would be:

$$\large (m + an)(m + bn) $$

It must be the case that $ a \cdot b = -2 $ and $ a + b = +1 $

Can you guess what a and b are from that?

larseighner · Answer

Well you have the second one.

anonymous · Answer

okay. how would i factor out the other one?

larseighner · Answer

As above.  What are a and b so that a x b = -2 and a + b = 1?