Linear DE's of higher order...

[T3achme_Isuck]

Not quite sure where to post this, as I'm a new member...
The DE is not the problem for me, it's FACTORING...LOL.

anywho, most ppl have said this won't factor, but my teach says there's a way.

m^3+m^2-2

I tried fractional powers (M^3/2), but it leaves me with an extra m^3/2, and I need m^2

Thanks for the help.

 

[galactus]

I'd use the Rational Root theorem.

In this case, we know a root is a factor of +/-2, +/-1.

Divide by (m-1) and see what happens.

\(\displaystyle \frac{m^{3}+m^{2}-2}{m-1}\)