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:

So the solutions of the cubic are .

Your problem can be solved in a similar manner.

RonL