The proofs of the fundamental Theorem of Algebra There are many proofs of the fundamental theorem of algebra.
Which are the most beautiful proofs?
 A: How about a proof in two parts:
$\bullet$ If $R$ is a real-closed field, then $R[X]/(X^2+1)$ is algebraically closed.
$\bullet$ $\mathbb R$ is a real-closed field.
A: Homotopy approach
Assume that a $p$ exists with $\deg p>0$ and $p(z)\neq 0$ for all $z\in\mathbb C$.
The homotopy approach is to note that a polynomial $p$ sends circles around zero with "large enough" radius to a path that goes around the origin $n$ times, counterclockwise, where  $n=\deg p$. But the circles of radius tiny radius go to paths that circle zero not at all, and these paths are trivially homotopic.
Then we just note that there is no homotopy in the punctured plane $\mathbb C\setminus 0$.
Of course, this requires a lot of machinery to prove rigorously, but I've always liked it because it makes intuitive sense, imaging trying to stretch a rubber band that does not "go around zero" to go around zero a total of at least once, without passing through zero...
This homotopy approach is much like a multi-dimensional "intermediate value theorem." If $f:\mathbb R^n\to\mathbb R^n$ sends a sphere to something that wraps around the origin in some non-trivial way (that is, a map that is not homotopically constant in $\mathbb R^n\setminus\{0\}$) then it must send some point on the interior of that sphere to the origin.
Constructive proofs exist
There exist entirely constructive proofs of FTA. In reverse mathematics, FTA is a result in $RCA_0$, which, roughly, models constructive number theory.
You can Google "constructive proof of fundamental theorem of algebra" to find specific examples.
A: In my opinion, the most beautiful proofs are the purely algebraic ones.  We need a little analysis, but only enough to show that every odd degree polynomial over $\mathbb{R}$ has a root, which follows from the Intermediate Value Theorem (we can't really do better than this, since we must use completeness of $\mathbb{R}$ in some way for the theorem to hold).
Suppose that $\mathbb{R}\subset F$ is a finite extension of fields.  If we can show that it has degree $\leq 2$, then the FTA follows.
Let $G$ be the Galois group $\operatorname{Gal}(F/\mathbb{R})$, and let $H$ be a Sylow $2$-subgroup of $G$.  $H$ has odd index, so the fixed field $F^H$ of $H$ has odd degree.  But a nontrivial extension of $\mathbb{R}$ of odd degree would give us an odd-degree polynomial with no root, so we have $F^H = \mathbb{R}\implies H=G$, so $G$ is a $2$-group.
If $|G|=1$ or $|G|=2$, we are done.  If $|G|\geq 4$, then $G$ has a subgroup of index $2$ containing a subgroup of index $4$ (by general facts about $p$-groups).  By Galois, these correspond to a degree $2$ extension of $\mathbb{C}$, which is impossible by the quadratic formula.
A: Consider $p: \mathbb R^2 \to \mathbb R^2$ a non-constant complex polynomial. Its Jacobian determinant for each $z \in \mathbb R^2$ is given by $\|p'(z)\|$, thus the determinant is zero at finitely many points, that is, the zeros of the polynomial $p'(z)$. Then $p(\mathbb R^2)$ is open (which is an exercise in analysis). On the other hand, $$\lim _{z \to \infty} p(z) = \infty$$
Then $p(\mathbb R^2)$ is closed. As $\mathbb R^2$ is connected we have that $p(\mathbb R^2) = \mathbb R^2$, that is , $p$ is surjective. That concludes the proof. 
A: This blog post has a probabilistic proof based on Brownian motion.
A: Via Galois Theory:
WLOG, let $p(x)\in\mathbb{R}[x]$ and let $F$ be the splitting field of $p$, embedded in some algebraic closure of $\mathbb{R}$. Since $F(i)$ is the splitting field of the square free part of $p(x)(x^2-1)$, it is Galois over $\mathbb{R}$. 
Let $G$ be the Galois group of $F(i)$ over $\mathbb{R}$. If there was an odd number dividing $|G|$, by Galois correspondence there is an odd degree subextension. This contradicts the fact that the intermediate value theorem tells us that every odd degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$, so it must be that $G$ has order $2^n$ for some $n$.
Let $G'$ be the Galois group of $F(i)$ over $\mathbb{C}$. This is the splitting field of some polynomial dividing $p(x)$ so it is Galois. By the above, $|G'|=2^k$, where we wish to show $k=0$. By Cauchy's Theorem for $p$-groups, if $k>0$ then there exists a subgroup of $G'$, and therefore a field extension of $\mathbb{C}$ that has order $2$, but that contradicts the quadratic equation, as the quadratic equation tells us that there are no degree $2$ irreducible polynomials over $\mathbb{C}$. Therefore $|G'|=1$. However, $2|G'|\geq |G|$ so $2\geq |G|$.
It follows by the Fundamental Theorem of Galois Theory that $[F(i):\mathbb{R}]\leq 2$. Since $$2=[F(i):\mathbb{R}]=[F(i):F][F:\mathbb{R}]$$ we have that $[F:\mathbb{R}]=2$ precisely when $F=\mathbb{C}$ and that $[F(i):F]=2$ precisely when $F=\mathbb{R}$
However, $F$ was an arbitrary finite degree field extension, so we have that an arbitrary finite degree field extension of $\mathbb{R}$ is $\mathbb{C}$ precisely when it is non trivial, and we are done.
A: See the book The Fundamental Theorem of Algebra by Fine and Rosenberger, which contains eight proofs.
See also The Fundamental Theorem of Algebra: A Visual Approach by Velleman.
A: Let $P\in\mathbb{C}[X]$ of degree greater than $1$. Assume by contradiction that $P$ has no roots in $\mathbb{C}$ and let define: $$f:=\frac{1}{P}.$$
$f$ is holomorphic on $\mathbb{C}$ and is bounded since: $$\lim_{z\to\pm\infty}|f(z)|=0.$$
Therefore, using Liouville's theorem, $f$ is constant and so is $P$, a contradiction.
