The Fundamental Theorem of Algebra tells us that any polynomial with real coefficients can be written as a product of linear factors over $\mathbb{C}$. If we don't want to use $\mathbb{C}$, the best we can say is:
If $p(x)$ is a polynomial with real coefficients, $\deg p > 2$, then there exists some quadratic polynomial $q(x)$ such that $q(x)$ is a factor of $p(x)$. (There are then two cases: either $q(x)$ is irreducible over $\mathbb{R}$, or it can be further factored as a product of linear factors.)
This is easily proven as a corollary of the FTA: If we work over $\mathbb{C}$, $p(x)$ can be written as a product of linear factors, and since a complex number $z$ is a root of $p(x)$ if and only if its conjugate $\bar{z}$ is, we can pair up linear factors so as to get a real quadratic.
(In fact, the theorem above -- which seems at first glance like a weaker form of the FTA -- is equivalent to it, since every real quadratic can be factored over $\mathbb{C}$.)
Is there a way to prove the above theorem without invoking complex numbers? It seems to have a certain value on its own as a property of the reals. For example, the fact that any even-degree real polynomial of $\deg 2n$ can be factored into $n$ (real) quadratics seems like something one should be able to prove without needing to change fields.
A version of this question was asked at Factorize real polynomials to quadratic factors. Proof without fundamental theorem of algebra., but the accepted (and only) answer there just concluded that such a proof would be equivalent to the FTA. I already know that; I'm wondering how one could write such a proof without passing to the algebraic closure.
Or, to put it another way: The statement of the theorem would make sense (and would be true) even if complex numbers had never been invented (or discovered, if you are a Platonist). So it seems like it ought to be provable without using complex numbers. Is it?
Here is a topological proof. It suffices to show that every irreducible polynomial over $\mathbb{R}$ has degree $\leq 2$, or equivalently that every finite field extension of $\mathbb{R}$ has degree $\leq 2$. So suppose $K$ is a finite extension of $\mathbb{R}$ of degree $d>2$. Note that then $K$ has a natural topology homeomorphic to $\mathbb{R}^d$ and the field operations are continuous, and so $K^\times\cong\mathbb{R}^d\setminus\{0\}$ is a topological abelian group.
We can now invoke some heavy machinery from topology to conclude this is impossible for $d>2$. For instance, any topological abelian group $X$ is weak homotopy equivalent to a product $\prod_n K(\pi_n(X),n)$ of Eilenberg-Mac Lane spaces (one for each of its homotopy groups). But $K(\pi_{d-1}(\mathbb{R}^d\setminus\{0\}),d-1)=K(\mathbb{Z},d-1)$ has nontrivial cohomology in infinitely many dimensions if $d>2$, and so $\mathbb{R}^d\setminus\{0\}$ cannot be weak equivalent to a product with $K(\mathbb{Z},d-1)$ as a factor.
Alternatively, you can observe that the topological abelian group structure on $\mathbb{R}^d\setminus\{0\}$ is actually smooth, so it makes $\mathbb{R}^d\setminus\{0\}$ an abelian Lie group. Any connected abelian Lie group is a product of copies of $\mathbb{R}$ and $S^1$, so this is impossible for $d>2$.