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proof of fundamental theorem of algebra (due to d’Alembert)
This proof, due to d’Alembert, relies on the following three facts:

Every polynomial with real coefficients which is of odd order has a real root. (This is a corollary of the intermediate value theorem.

Every second order polynomial with complex coefficients has two complex roots.

For every polynomial $p$ with real coefficients, there exists a field $E$ in which the polynomial may be factored into linear terms. (For more information, see the entry “splitting field”.)
Note that it suffices to prove that every polynomial with real coefficients has a complex root. Given a polynomial with complex coefficients, one can construct a polynomial with real coefficients by multiplying the polynomial by its complex conjugate. Any root of the resulting polynomial will either be a root of the original polynomial or the complex conjugate of a root.
The proof proceeds by induction. Write the degree of the polynomial as $2^{n}(2m+1)$. If $n=0$, then we know that it must have a real root. Next, assume that we already have shown that the fundamental theorem of algebra holds whenver $n<N$. We shall show that any polynomial of degree $2^{N}(2m+1)$ has a complex root if a certain other polynomial of order $2^{{N1}}(2m^{{\prime}}+1)$ has a root. By our hypothesis, the other polynomial does have a root, hence so does the original polynomial. Hence, by induction on $n$, every polynomial with real coefficients has a complex root.
Let $p$ be a polynomial of order $d=2^{N}(2m+1)$ with real coefficients. Let its factorization over the extension field $E$ be
$p(x)=(xr_{1})(xr_{2})\cdots(xr_{d})$ 
Next construct the $d(d1)/2=1$ polynomials
$q_{k}(x)=\prod_{{i<j}}(xr_{i}r_{j}kr_{i}r_{j})$ 
where $k$ is an integer between $1$ and $d(d1)/2=1$. Upon expanding the product and collecting terms, the coefficient of each power of $x$ is a symmetric function of the roots $r_{i}$. Hence it can be expressed in terms of the coefficients of $p$, so the coefficients of $q_{k}$ will all be real.
Note that the order of each $q_{k}$ is $d(d1)/2=2^{{N1}}(2m+1)(2^{N}(2m+1)1)$. Hence, by the induction hypothesis, each $q_{k}$ must have a complex root. By construction, each root of $q_{k}$ can be expressed as $r_{i}+r_{j}+kr_{i}r_{j}$ for some choice of integers $i$ and $j$. By the pigeonhole principle, there must exist integers $i,j,k_{1},k_{2}$ such that both
$u=r_{i}+r_{j}+k_{1}r_{i}r_{j}$ 
and
$v=r_{i}+r_{j}+k_{2}r_{i}r_{j}$ 
are complex. But then $r_{i}$ and $r_{j}$ must be complex as well. because they are roots of the polynomial
$x^{2}+bx+c$ 
where
$b={k_{2}u+k_{1}v\over(k_{1}+k_{2})}$ 
and
$c={uv\over k_{1}k_{2}}$ 
Note. D’Alembert was an avid supporter (in fact, the coeditor) of the famous French philosophical encyclopaedia. Therefore it is a fitting tribute to have his proof appear in the web pages of this encyclopaedia.
References
 1 Jean le Rond D’Alembert: “Recherches sur le calcul intégral”. Histoire de l’Acadḿie Royale des Sciences et Belles Lettres, année MDCCXLVI, 182–224. Berlin (1746).
 2 R. Argand: “Réflexions sur la nouvelle théorie d’analyse”. Annales de mathématiques 5, 197–209 (1814).
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Comments
D'Alembert's proof
I read about this wonderful proof in the book "Numbers" by Ebbinghaus et al. Since it's been a number of years since I lost my copy of this book, would someone out there who has access to a copy be kind enough to provide me with bibliographis information that I could add as a reference. Even better, maybe someone could provide a reference to the original publication by D'Alembert.
Re: D'Alembert's proof
I think I have a copy at home. If no one else responds, I'll try to remember to send you something tonight (though a PM mail couldn't hurt to remind me)
Thanks,
Cam
Re: D'Alembert's proof
Apparently, the proof was only really made rigorous by R. Argand in "Reflesions sur la nouvelle theorie d'analyse." (Annales de mathematiques 5, 197209, 1814), but the idea of the proof can indeed be traced back to D'Alembert.
D'Alembert's proof, which Gauss said was not entirely satisfactory from a rigour point of view, can be found in "Recherches sue le calcul integral," in (Histoire de l'Academie Royale des Sciences et Belles Lettres), annee MDCCXLVI, Berlin 1748, 182224)
Hope this helps,
Cam
Re: D'Alembert's proof
Thank you for the references. I have added them to the entry.
The objection which Gauss and Argand raised was that D'Alembert's proof rests on the assumption that any polynomial has roots. Today, we understand that this is correct since, given a polynomial over a field, one can always construct an extension of the field in which the polynomial factors into linear factors. At the turn of the ninteteenth cenrury, the situation was a lot different. Mathematicians were still struggling to show that the complex number system was consistent (Argand planar representation of complex numbers went a good way towards reassuring mathematicians that complex numbers were legitimate) and abstract algebra was in a very rudimentary state. A good example of this is the fact that even the grandmaster Gauss had very little conception of group theory. In his disquisitions, Gauss demonstrates certain propositions about the composition of quadratic forms. Today, we would recognize them as the statement that quadratic forms form an Abelian group under composition, but, judging from the confused manner in which he explains what should be the statement of associativity, Gauss was not familiar with even the rudiments of group theory which any freshman of today would know.
It wasn't until the latter part of the ninetennth century that the necessary concepts of vector spaces, fields and the like were in place that Kronecker was able to show that splitting fields exist. This answered the objection to D'Alembert's proof (the proof of splitting fields in no way assumes the fundamental theorem of algebra, so there is no queestion of circularity) and we today can accept his proof as correct.
I think I will add a polished version of thease remarks as a historical postcript to my entry on D'Alembert's proof.
Thanks again,
Ray