Stanton D. answered • 04/23/14

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Theresa,

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)