Getting to the Root of the (Polynomial) Matter

You are often asked to find all the zeros (roots or x-intercepts) of polynomials.

To do this in the most efficient way, use the rational zero test and your graphing calculator.

First, there are some general concepts.

When you FOIL a pair of quadratic binomials with leading coefficients of one, notice that for , the constant in the trinomial function is the product (r₁ r₂).
When we to do the multiplication for , the constant in the polynomial function is the product (r₁ r₂ r₃).
That means that all rational zeros must be in the factor list for the constant in a polynomial function when the leading coefficient is one.

Next, remember that for the quadratic , the zeros are calculated using

So the leading coefficient a gets into the zero finding game as a divisor. This makes sense in general because to get from to the place where we can use the idea that the rational roots are factors in the constant, we would need to factor out a.

Then . And, .

In general, for a polynomial with integer coefficients,

where a is the leading coefficient, c is the constant in the polynomial, and the r_i are all the distinct roots repeated as often as needed. We reduce the fraction to simplify our work so the fraction indicates the reduced ratio.

We're almost ready to learn how to find zeros, but first we handle the bad news.

The r_i include the rational roots, irrational roots as irrational conjugate pairs , and the complex zeros as the complex conjugate pairs .

The only ones we are likely to find graphically are the rational zeros. If we don't know how to set up the calculator to help us, we might not even find those!

Here's how we do it:

The constant is the largest possible rational zero.
(Why is this true?)

So, if we use the window where x-min is and x-max is , all rational zeros must be in the window.

While we should "see" all the zeros on the graph, we might confuse rational zeros like x = 2 with an irrational zero like . They are too close together to differentiate easily by graph.

We need a table! If we set the table delta or table pitch to be , we should be able to track them all down.

It is really easy. Let's look at an example:

Finding Rational Zeros

Find all zeros of .

The Rational Zero Test tells us that the possible rational zeros are .

We could test those eight possibilities using synthetic division. However, I am way too lazy for that!
Using your calculator do the following (according to your calculator's method):

You want to reproduce the windows to below.

Notice I cheated a little to create the example. The middle window has the factored form of our function. Multiply out Y1 below to see if you get the correct polynomial. Remember just because we write it doesn't make it so!

Here are the steps:

Type in the equation.
Set a table to start at x = -5 or set the table x range of [-5, 5].
Set a table delta (pitch) of . The calculator will approximate in the display. But it uses so precise a calculation to build the table, that we can use the results.
Look at the table. You may have to scroll to see all the good stuff.

Piecing together the table data: The rational zeros are at x = -5,1, and by a reasonable interpretation, at .

It is reasonable because we set the pitch at one third. The dumb machine won't increment any other way.

Also, the value of y is -7.466666667x10^-13. My poor calculator is doing its best to say zero for output!

Also, we can substitute the value into the original polynomial and get exactly zero for y.

Since the polynomial is cubic, no other zeros exist.

To prove all of the zeros I can test each with synthetic division.
Do this smartly! Always check your integer possibilities first. Then the fractional values and other more ugly stuff will probably pop up at the end.

Well, wow! We don't need that synthetic division, right?

Wrong!

Synthetic division is the best mechanism for finding the irrational and complex zeros. The next example get's that point across:

Finding Irrational Zeros

Find all zeros of . Notice "all" means rational, irrational and complex.

Again, the Rational Zero test tells us that the possible rational zeros are .

The general setup is just like before, but we will get a different result. Here are the tables. Try to reproduce them with your calculator.

No matter how you look, you won't find the other two zeros by table. (Let's discount blind luck, please!)

Look at the graph. You should always refer back to it!

zooming to

All the action is between -5 and 5.

We see the zeros at x = -5 and 1 (barely).

When we zoom in on the area around the origin, we can see that there are two more real zeros between -1 and 1.

Now we bring out the heavy guns. With the two known zeros we use synthetic division to find the rest.

The synthetic division is shown to the left. We don't restart the process at each step. We just divide until we run out of rational zeros.

The last line tells us that we have an irreducible quadratic, .

So we solve it by the quadratic formula or any other exact method to get two irrational zeros, .

Finding Complex Zeros

Look at this example. The polynomial and synthetic division are in the box.

Graph the polynomial to confirm the rational zeros. Then click on the polynomial to see the graph. Notice that the graph failed to reach the x-axis. The complex zeros have created a "wiggle" in the graph.

The set up is just like before. However, now we have discovered an irreducible quadratic with complex zeros.

The irreducible quadratic solves to give us zeros at x = .

Using the rational root theorem and the graphing calculator together is a powerful technique. It does fail occasionally. The possibility of both irrational and complex zeros for the polynomial cannot be ignored. These are tough nuts to crack.

When that happens, break out a bigger hammer! We revert to programs such as MathCad^®, or Maple^®, or Mathematica^®. These programs can handle these types of polynomials, but require a little more experience and a computer. They also are not allowed during testing!

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