Find all the
zeroes of 6x^{4} – 13x^{3} – 34x^{2} + 80x – 25.

The Rational Roots Test
says that the possible zeroes are:

It is easy to check if x = –1 or x = 1 is a zero: plug
them in! You will be multiplying the coefficients by –1 or by 1,
according to the x-value
and the exponent, which is simple enough:

f(–1)
= 6 + 13 – 34 – 80 – 25 = –120

f(1)
= 6 – 13 – 34 + 80 – 25 = 14

So neither x = –1 nor x = 1 is a zero, but,
because of the sign change between f(–1)
= –120 and f(1) = 14,
there must be a zero in between x = –1 and x = 1. However, I'd
rather avoid fractions if at all possible, so I'm going to try the next
biggest non-fractional possible zeroes, being x = –5 and x = 5.

First I'll try x = –5:

So x = –5 is not a zero.
However, f (–5) = 4100 is positive
and f (–1) = –120 is negative,
so there must be a zero in between x = –5 and x = –1. And since f (–1) = –120 is "closer"
to zero than is f (–5) = 4100, I would
guess that the zero is "closer" to x = –1 than to x = –5. Also, I can
see that there is no need to test zeroes below x = –5, because I divided
by a negative, and the bottom row has alternating signs.

Let's see if x = 5 is a zero:

So x = 5 is not a zero.
Also, f(5)
= 1650 and f(1)
= 14 are positive,
so I can't tell if there is a zero in between x = 1 and x = 5.

Descartes' Rule
of Signs says that f(x)
= 6x^{4} – 13x^{3} – 34x^{2} + 80x – 25 has
three or one positive zeroes (so we are assured of at least one positive
zero!) and one negative zero. So there are either two real-number zeroes
(so the other solutions of this fourth-degree polynomial are complex)
or four real-number zeroes. Since there is only one negative root and
possibly as many as three positive roots, it looks to me as though my
odds of finding a rational zero are better if I try the positive side.

Since f(–1)
= –120 is negative
and f(1)
= 14 is positive,
then there must be a zero between x = –1 and x = 1. I'll try x = ^{1}/_{2}:

Nope. But the remainder
says that f(^{ 1}/_{2}) = ^{21}/_{4},
which is a positive number. So I know, from the sign change, that there
must be a zero between x = –1 and x = ^{1}/_{2}.
I'll try x = ^{1}/_{3}:

This isn't a root either.
Also, since the remainder says that f(^{ 1}/_{3}) = ^{–68}/_{27},
the zero must be between x = ^{1}/_{3} and x = ^{1}/_{2}.
However, there are no rational candidates between x = ^{1}/_{3} and x = ^{1}/_{2},
so the zero must be irrational. I'll have to look elsewhere for a rational
zero.

I have better odds of
finding positive zeroes than negative ones (from Descartes' Rule of
Signs), so I will try to find a zero between x = 1 and x = 5. I'll try x = ^{5}/_{2}:

I still have to find
at least one more zero, so I can reduce the polynomial to a quadratic,
but now I only have to find the zeroes of the result of the last division, 6x^{3} – 3x^{2} – 39x + 15.
That is, I am trying to solve 6x^{3} – 3x^{2} – 39x + 15 = 0.
Why not make life easier on myself, since everything happens to be a
multiple of 3,
and divide off the 3?
This leaves me with 2x^{3} – x^{2} – 13x + 5 = 0.
This also reduces the number of possible rational zeroes; my list now
consists only of:

I have already eliminated –5, 5,
–1, 1, ^{–1}/_{2}, and ^{1}/_{2} from my list of possible zeroes. Combining these earlier deletions with
my new shorter list leaves me with only ^{–5}/_{2} and ^{5}/_{2} to try. This polynomial has been messy so far, so I'll try the negative
value: x = ^{–5}/_{2}:

And there's my other
rational zero. Finally!

Now I'm left with just
a simple quadratic, which I can solve with the Quadratic
Formula:

2x^{2} – 6x + 2 = 0

x^{2} – 3x + 1 = 0

Then the zeroes are:

Phew! That was painful!
But be assured: I did not "go out of my way" to make these examples
difficult. They are actually fairly representative, and the steps shown
in these examples display the path I actually took (with pencil and paper)
to try to find the zeroes of these polynomial. There is no magic "trick"
to these exercises, there is no algorithm, there is no "secret".
They just take time. So give yourself some time, grab some scratch paper,
and practice, practice, practice!

Advisory: In "real
life", there is absolutely no guarantee that a given polynomial will
have a rational root. As a matter of fact, there probably won't be. In
these cases, using graphing software or some other numerical
method is the only
practical way to approximate the zeroes. However, in an algebra class,
when they give you this type of problem, you can generally rest assured
that there will be enough rational roots to let you get down to a quadratic.
(That is, when they tell you to find the zeroes, they will have given
you zeroes that are actually find-able.) You should not be surprised,
however, when the final quadratic spits out irrational or complex-valued
stink-nasties.