This is devoted to a mostly algebraic proof of the fundamental theorem of algebra. It depends on the interesting results about symmetric polynomials which are presented above. I found it on the Wikipedia article about the fundamental theorem of algebra. You google “fundamental theorem of algebra” and go to the Wikipedia article. It gives several other proofs in addition to this one. According to this article, the first completely correct proof of this major theorem is due to Argand in 1806. Gauss and others did it earlier but their arguments had gaps in them.
You can’t completely escape analysis when you prove this theorem. The necessary analysis is in the following lemma.
Proof: This follows from the intermediate value theorem from calculus.
Next is an algebraic consideration. First recall some notation.

Recall a polynomial in
The following is the main part of the theorem. In fact this is one version of the fundamental theorem of algebra which people studied earlier in the 1700’s.
Proof: It is possible to write

where m is odd. If n is odd, k = 0. If n is even, keep dividing by 2 until you are left with an odd number. If k = 0 so that n is odd, it follows from Lemma 10.3.1 that p
 (10.15) 
where p_{k}
 (10.16) 
There is another polynomial which has coefficients which are sums of real numbers times the p_{k} raised to various powers and it is

I need to verify this is really the case for q_{t}
Note that the degree of q_{t}


and so by induction, for each t ∈ ℝ,q_{t}
There must exist s≠t such that for a single pair of indices i,j, with i < j,

are both complex. Here is why. Let A
Now for that t,s,

and also

At this point, note that z_{i},z_{j} are both solutions to the equation

which from the above has complex coefficients. By the quadratic formula the z_{i},z_{j} are both complex. Thus the original polynomial has a complex root. ■
With this lemma, it is easy to prove the fundamental theorem of algebra. The difference between the lemma and this theorem is that in the theorem, the coefficients are only assumed to be complex. What this means is that if you have any polynomial with complex coefficients it has a complex root and so it is not irreducible. Hence the field extension is the same field. Another way to say this is that for every complex polynomial there exists a factorization into linear factors or in other words a splitting field for a complex polynomial is the field of complex numbers.

Proof: First suppose a_{n} = 1. Consider the polynomial q

and the sum involves adding terms of the form

so it is of the form of a complex number added to its conjugate. Hence q
Next suppose a_{n}≠0. Then simply divide by it and get a polynomial in which a_{n} = 1. Denote this modified polynomial as q

where q_{1}

and continue this way. Thus

Why use this more elaborate proof? I think it is because you can give other examples of algebraically complete fields. For example, begin with ℚ and let the algebraic numbers be denoted by A. These are those numbers which are roots of a polynomial having rational coefficients. Then consider A^{2} to be those complex numbers which are roots of a polynomial having coefficients in A. In general, let A^{n} be roots of polynomials with coefficients in A^{n−1}. In general, if A^{n−1} is countable, then so is A^{n}. This is routine to show using the fact that there are countably many polynomials of degree m for each m ∈ ℕ. Each has at most m roots. Thus A^{∞} ≡∪_{n=1}^{∞}A^{n} is countable because ℚ is. Now recall also that it was shown that the algebraic numbers over a field are a field. Therefore, A^{∞} is also a field because any finite number of elements of A^{∞} must be in a single one of the fields A^{n} for large enough n. Now consider Lemma 10.3.1 applied to a polynomial having real coefficients in A^{∞}. These coefficients are in some A^{n} and so the root from ℂ having these coefficients is in A^{n+1} ⊆ A^{∞}. Now the rest of the argument goes similarly. You show using the same considerations that every polynomial having real coefficients in A^{∞} has a root in A^{∞}. Then you do the easy extension to the case where the coefficients in A^{∞} are complex. This field is clearly much smaller than ℂ because it is countable, and yet it is algebraically complete. The standard analysis proof given earlier will obviously not work because it is based on compactness considerations.