he general technique for solving bigger-than-quadratic polynomials is
pretty straightforward, but the process can be time-consuming.

The first step is to apply the Rational Roots Test to the polynomial
to get a list of values that might possibly be solutions to the
polynomial equation. You can follow this up with an application of
Descartes' Rule of Signs, if you like, to narrow down which possible
zeroes might be best to check. Of course, if you've got a graphing
calculator, it's a good idea to do a quick graph, since x-intercepts
of the graph are the same as zeroes of the equation. Seeing where the
graph looks like it crosses the axis can quickly narrow down your list
of possible zeroes.

Once you've found a value you want to test, you use synthetic division
to see if you can get a zero remainder. If you get a zero remainder,
you've not only found a zero, but you've also reduced your polynomial
by one degree. Remember that synthetic division is, among other
things, division, so checking if x = a is a solution is the same as
dividing out the linear factor x – a. This means that you should not
return to the original polynomial for your next computation (for
finding the other zeroes); you should instead work with the output of
the synthetic division. It's smaller, so it's easier to work with.

You should not be surprised to see some complicated solutions to your
polynomials (that is, solutions containing square roots or complex
numbers, or both); these zeroes will come from applying the Quadratic
Formula to the last (quadratic) factor of your polynomial. Here's how
the process plays out:

* Find all the zeroes of the following polynomial: 2x5 + 3x4 –
30x3 – 57x2 – 2x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is check to see if x = 1
or x = –1 is a root, because these are the simplest roots to test for.
This isn't an "official" first step, but it can often be a timesaver,
because you can just look at the powers and the numbers. When x = 1,
the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1
isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so
x = –1 is a root, and I can take care of it right away:

synthetic division by x = -1

This leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x
+ 24. (Since I've divided out the factor x + 1, I've reduced the
degree of the polynomial by 1. That's how I know this is a degree-four
polynomial.) Now I'll apply the Rational Roots Test to get a list of
potential zeroes to try:

±1/2, 1, 3/2, 2, 3, 4, 6, 8, 12, 24

From experience, I've learned that most of these exercises have
their zeroes near the middle of the list, rather than at the extremes.
This isn't always true, of course, but it's usually better to stay
away from the larger numbers. In this case, I won't start off by
trying stuff like x = –24 or x = 12. Instead, I'll start out with
smaller values like x = 2. And I can narrow down my options further by
"cheating" and looking at the graph:

graph of y = 2x^4 + x^3 - 31x^2 - 26x + 24

This is a fourth-degree polynomial, so it has, at most, four
x-intercepts, and I can see all four of them on the graph. It looks
like one of the zeroes is around –3.5, but –7/2 isn't on the list that
the Rational Roots Test gave me, so this must be an irrational root.
I'll leave it until last.

It also looks like there may be zeroes near –1.5 and 0.5. But
the clearest solution looks to be at x = 4 and since whole numbers are
easier to work with than fractions, x = 4 would probably be a good
value to try:

synthetic division with x = 4

The zero remainder says that x = 4 is a root. The bottom row of
the synthetic division tells me that I'm now left with factoring 2x3 +
9x2 + 5x – 6. Looking at the constant term "6", I can see that x =
±24, ±12, ±8, and –4 won't work as rational roots (even if I didn't
already know from the graph), so I can cross them off of my list.
(Always check the numbers as you go. The Rational Roots Test can give
a very long list of possibilities, and it can be helpful to notice
that some of those values can be ignored, especially if you don't have
a graphing calculator to "cheat" with.) Comparing the remaining values
on the list with the intercepts on the graph, I'll try x = 1/2:

synthetic division with x = 1/2

The remainder isn't zero, so that test root didn't work. This
means that the zero close to x = 1/2 on the graph must be irrational;
I'll find it when I apply the Quadratic Formula later. For now, I'll
try x = –3/2: Copyright © Elizabeth Stapel 2005-2011 All Rights
Reserved

synthetic division with x = -3/2

The division came out evenly, leaving me with the polynomial 2x2
+ 6x – 4. Since I'm looking for the zeroes of the polynomial, what I
really have here is 2x2 + 6x – 4 = 0. Dividing through by 2 to get
smaller numbers gives me x2 + 3x – 2 = 0, to which I can apply the
Quadratic Formula:

x = [-3 ± sqrt(17)]/2, or about -3.5 and 0.5

Then the complete solution is:

x = -1, [-3 +/- sqrt(17)]/2, -3/2, 4

Asking you to find the zeroes of a polynomial means the same thing as
asking you to find the solutions to a polynomial equation. The zeroes
of a polynomial are the values of x that make the polynomial equal to
zero. So the above problem could have been stated along the lines of
"Find the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 = 0" or "Find
the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x = –24", and the answers
would have been the exact same list of x-values.

You can use these same techniques to factor bigger-than-quadratic
polynomials....