Fundamental theorem of algebra applied to $y=x^2$ The fundamental theorem of algebra seems to indicate that there should be two roots for $y=x^2;$ however, applying the quadratic formula leads to just $0.$ Considering that $+0$ and $-0$ make no sense, what is the reasoning behind this superficial contradiction?
 A: The issue that has attracted your interest is a statement of the fundamental theorem of algebra that reads roughly as follows: "A complex polynomial $p$ of $\deg n$ has $n$ roots in $\newcommand{\CC}{\mathbb{C}}\CC$."
You've then found a polynomial for which this is apparently violated,
$x^2$. Essentially the issue is that since $x^2=(x-0)(x-0)$, there is only one root, but it's repeated twice. There are two more or less equivalent ways to address this apparent discrepancy. (1) Clarify what we mean by "$n$ roots," and (2) give a restatement of the FTA which removes the discrepancy in another manner.
I'll first give a restatement of the FTA for which there isn't any ambiguity: Any complex polynomial $p$ of degree $n$ factors as a product of $n$ linear factors: $$p(z)=\prod_{i=1}^n (a_iz-b_i),$$
for $a_i,b_i\in\CC$, $a_i\ne 0$. Now that we aren't counting roots, but rather linear factors, there is no ambiguity when we have a repeated factor. As we can see your example $z^2=z\cdot z$ factors just fine into two linear factors.
Now I'll address the other option. When we say $p$ has $n$ roots, we mean $p$ has $n$ roots counted with multiplicity. We say $p$ has a root of multiplicity $n$ at $a\in\CC$ if $n$ is the largest integer such that $(z-a)^n\mid p(z)$. I.e. the multiplicity of a root $a$ is the number of times the corresponding linear factor $(z-a)$ divides $p$. Let $m_p(a)$ denote the multiplicity of $p$ at $a$ (note that if $p(a)\ne 0$, then we can define $m_p(a)=0$, so we can extend the definition of multiplicity to nonroots as well, since then $m_p(a) > 0$ if and only if $a$ is a root of $p$). Then the fundamental theorem of algebra says more precisely that 
$$\deg p = \sum_{a\in Z(p)} m_p(a) = \sum_{a\in\CC} m_p(a),$$
where $Z(p) = \{a\in\CC : p(a) = 0 \}$ is the set of zeroes of $p$.
In this case $\deg z^2 = 2 = m_{z^2}(0)$.
A: I don't know what exact statement of the Fundamental Theorem of Algebra you're using. The Fundamental Theorem of Algebra guarantees at least one root. Sometimes you'll see it stated that for a polynomial degree $n$, there are $n$ roots (some with multiplicity), but multiplicity is defined precisely by how many times the factor $(x-\lambda)$ is repeated in factorization of the polynomial. In this case, $0$ is a root of multiplicity $2$ because $x^2 = (x-0)(x-0)$.
A: The Fundamental Theorem of Algebra allows multiple roots. In this case$x=0$ counts twices so you have a root of multiplicity two. For example $y=(x-1)^3(x+3)^2$ which is a polynomial of degree $5$,has $x=1$ with multiplicity $3$ and $x=-3$ with multiplicity $2$. 
