**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?

**exactly**and

**only**the same as the zeros of each of the terms (factors) you just grouped the function into.

**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.

**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.