Lets look at the very similar equation:
First thing to do is sketch the graph of:
to see if we can find the approximate location of its zeros.
From the sketch in the attachment we see that this has zeros
near -2.7, -2, and +2.7. Trying these out we find that -2 is
an exact root of the cubic. So we may write:
Now you can use synthetic division to find the quadratic, which
in this case turns out to be:
which may be factorised either by inspection of by using the
quadratic formula to find its roots, and hence its linear factors.
So the solutions of the cubic are .
Your problem can be solved in a similar manner.