Does $x^{-1}$ have -1 zeroes? I asked How $y=x^4+1$ could have 4 zeroes yesterday, and figured it out on my own pretty quickly. I did some more thinking since then, and realized that according to the fundamental theorem of algebra, $0=x^{-1}$ has -1 zeroes (solutions?), because $-1$ is the largest exponent.
Wouldn't it have $1$ zeroes though, because $\sqrt[-1]{0} = 0^{\frac{1}{-1}} = 0$? Or does it equal undefined? Or something else? Is there something important I'm missing, or does the question not make sense at all?
 A: This is a great question!  As rschweib explained, you cannot literally apply the fundamental theorem of algebra to things which aren't polynomials.  However, there is a souped-up version which applies to any ratio of two polynomials, such as $x^{-1}=1/x$.  
Say you are interested in the function $f(x)=p(x)/q(x)$, where, for simplicity, let's suppose we wrote $p$ and $q$ without common factors.  The zeros of $p$ are zeros of $f$, while the zeros of $q$ are places where $f$ is singular.  We say $f$ has a "pole" there.
The formula is very simple: for any rational function, the number of zeros, counted with multiplicity, equals the number of poles, counted with multiplicity.  
If you are paying attention, though, you will see that that doesn't seem to agree with our experience, since polynomials have lots of zeros and no poles -- or so it appears!
To make the formula work out correctly, you should also count infinity as a possible zero or pole.  The order of the pole of $f$ at infinity is the degree of $p$ minus the degree of $q$.  If the order is negative, then it counts as negative that many zeros.  
So $1/x$, for example, has a pole of order 1 at 0, and a pole of order -1 (i.e. a zero of order 1) at infinity.  And 1=1, as predicted.  
Calling the behaviour at infinity a zero or a pole really does make pretty good sense: in the example of $1/x$, you can see that as $x$ goes to infinity, $f(x)$ really does approach zero, while for a polynomial, as $x$ goes to infinity, $f(x)$ becomes very large in absolute value, so it makes sense to think of it as a pole.  
A: The fundamental theorem of algebra deals with polynomials only (elements of $\mathbb C[x]$, usually), and $x^{-1}$ is not a polynomial.
$x^{-1}$ is an element of the field of fractions of $\mathbb C[x]$, but it is not properly considered a polynomial itself. It can be called a Laurent polynomial, however.
Trying to use the fundamental theorem of algebra with Laurent polynomials is unproductive. For example, $x^{-1}(x-1)(x-2)(x-3)$ has "degree" $2$ when you multiply it out, and yet it clearly has three roots.
A: Let us consider the roots of a generalized polynomial equation such as
$$ax^3+bx^2+cx+d+ex^{-1}+fx^{-2}=0.$$
$x=0$ does not belong to the domain, so that we can multiply by $x^2$ and get an ordinary polynomial equation:
$$ax^5+bx^4+cx^3+dx^2+ex+f=0$$
which has five roots.
For such "polynomials", the number of roots is the algebraic difference of the highest and the lowest (negative) degrees ($3-(-2)=5$).

This reasoning also works with $x^{-1}$, as the difference between the extreme degrees is $0$, showing that no root is possible ($x^{-1}=0$ is turned into $1=0$).
