Clarification on "Every polynomial function of degree $\ge1$ has at least $1$ zero in the complex number system."

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.

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.