x^3 + x^2 - 9x - 9 = 0
Now first note that the degree of this equation is 3. That is, the highest power of x is 3. That means that there are 3 zeros/roots, but we don't know whether these roots are real numbers or imaginary.
Before trying to find the roots we can first find out the number of possible real roots by using something called "Descartes rule of signs". Look this up (it'll be easier to understand if you just see examples of it).
Using Descartes rule of signs on our new equation (x^3 + x^2 - 9x - 9 = 0) for f(+x) we have ++--, which means there is only one sign change. Therefore there is either 1 or 0 positive real roots. For f(-x) we have -++- which means there are two sign changes. Therefore there is either 2 or 0 negative real roots.
Now we can first look for rational roots by using something called the "rational root theorem". Again look this up. The rational root theorem tells us that the possible rational roots of this equation are:
+-9/1, 3/1, 1/1
We can see if any of these are roots by either using "synthetic division" or by plugging the values into the equation. By letting x=-1 you'll see that it is a root.
Since x=-1 is a root, we can rewrite our equation as (here is where you either have to eyeball the equation or use synthetic division):
(x+1)(x^2 - 9) = 0
The (x^2 - 9) term can be written as (x+3)(x-3) so our equation is now:
(x+1)(x+3)(x-3) = 0
We now know that x = +-3 are roots, and since from earlier we know that there are only 3 roots we're done:
x = {-3, -1, 1}
NOTE: Many times your equation won't have all rational roots, so you'll have to use the "quadratic formula" to find the rest of the roots. I'll cite a useful site.