In order to answer questions like this, you need always to factor your polynomial expression. Fortunately, this one is simple, and mostly done for you. The x^3 is already packaged up. The (x^2-81) needs a bit more work: since it's the difference of two squares (x^2 and 81=9^2) there's a special factoring that you should by now be able to do in your sleep: (x+9)(x-9) right?
So, you write: f(x) = (x^3)(x+9)(x-9)
OK, now you have your terms ready to go. The zeros of the overall function are exactly and only the same as the zeros of each of the terms (factors) you just grouped the function into.
So solve each of the factors for zero:
x^3=0 --> true only for x=0, right? Anything else, when you cube it, you'll get a number other than zero!
For this factor, because it's a cube, the multiplicity is 3 (technically, that's like having a solution of zero, three separate times).
x+9=0 --> true for x=-9 (right?), multiplicity 1;
x-9=0 --> true for x=9 (right?), multiplicity 1.
One other thing you should know: if you graph this function (on a calculator, or by hand) the zeros of the function (the places where it touches or crosses the x-axis) are exactly at the x-values you just solved to get.
OK, you say, that's neat, but so what? It turns out that most of the time when we solve a problem
of any kind involving numbers we're actually done the equivalent of finding the zeros for some function. So, if I say, "My speed squared is 4. What is my speed?" I've solved the problem x^2=4 which is the same as x^2-4=0. That looks like a trivial distinction to you (maybe), but rest assured that in general, the x^2-4=0 form is much easier to solve than the x^2=4 form for more difficult problems. So it's a great idea to get as familiar with functions and their zeros as you can.
Hope this helps you a bit, Theresa.
--S. (just down the road from you in Hamilton, NJ, actually)