Clarification on "Every polynomial function of degree $\ge1$ has at least $1$ zero in the complex number system." The Fundamental Theorem of Algebra says "Every polynomial function of degree $\ge1$ has at least $1$ zero in the complex number system." 
My question is, where do the rest of the zeroes of the polynomial lie? Can it happen that they do not belong to the complex number system? Would we then have to pass to a number system beyond the complex numbers? 
 A: The theorem may as well state that every polynomial equation of degree $n$ has exactly $n$ roots (counted with their multiplicity). The statements are equivalent, for, if your polynomial $p(z)$ of degree $n$ has one root $\lambda$, then you can factor it as
$$
p(z) = (z-\lambda)q(z),
$$
where the degree of $q$ is $n-1$ Then you can recursively apply the result to $q$, until you reach a polynomial of degree 1.
A: This is a statement of the Fundamental Theorem of Algebra that is completely correct, but in a way kind of obscures what is actually going on if someone doesn't think about what this actually means...
Consider a complex polynomial $p(z) = a_0 + a_1z + ... + a_nz^n$. By the fundamental theorem of algebra, this has at least 1 zero. It is a well-known result that when a polynomial has a zero at $r$ we can write:
$$p(z) = (z-r)(b_0 + ... + b_{n-1}z^{n-1}) = (z-r)(q(z))$$
for some polynomial $q(z)$ of degree $n-1$. 
By the fundamental theorem of algebra we have that q(z) (as long as it is not constant) has at least 1 zero, so we factor that zero out and repeat until the degree of the not-yet-factored part (in this case, $q(z)$) is constant.
An induction argument can make this far more formal.
The confusion ensues because the zero of $q(z)$ can be the same as the zero of $p(z)$. So we could have that a degree $n$ polynomial $p(z)$ has only one (distinct) zero. For example $p(z) = (z+1)^n$. This polynomial only has 1 distinct zero, but it has multiplicty $n$.
But we can use the argument I had outlined above to phrase the fundamental theorem of algebra in a different way (or I've heard some people claim this is a corollary):
A nonconstant polynomial of degree $n$ has, counting multiplicity, exactly $n$ zeros.
