Proving (without using complex numbers) that a real polynomial has a quadratic factor

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 $d>2$.