Most even-degree polynomials in $\mathbb{R}[x]$ are reducible Say $f(x) \in \mathbb{R}[x]$ has degree $n \geq 3$. If $n$ is odd, then $f(x) \to + \infty$ as $x \to +\infty$ and $f(x) \to - \infty $ as $x \to -\infty$ (or the other way around), so $\exists c \in \mathbb{R}$ such that $f(c) = 0$ by the intermediate value theorem, so $f(x) = (x - c)g(x)$ for a nonconstant polynomial $g(x) \in \mathbb{R}[x]$, so $f(x)$ is reducible.
The same is true if $n$ is even: $f(x)$ is still reducible. Can we prove this without using the fact that $\mathbb{C}$ is algebraically closed? Is there any intuitive reason that an even-degree polynomial in $\mathbb{R}[x]$ is reducible if it is not quadratic? The preceding paragraph gives a simple explanation for why odd-degree polynomials behave this way.

Here is a proof using algebraic closure. First, we must prove that $\mathbb{C} \cong \mathbb{R}[x]/(x^2 + 1)$ is indeed algebraically closed. Let $f \in \mathbb{C}[x] - \mathbb{C}$, and take $M$ to be the splitting field of $f$ over $\mathbb{C}$, then take $N$ to be a normal extension of $\mathbb{R}$ containing $M$, so that $\mathbb{R} \subset \mathbb{C} \subseteq M \subseteq N$; it suffices to prove that $N = \mathbb{C}$. Let $H$ be a Sylow 2-subgroup of $\mathrm{Gal}(N/\mathbb{R})$, so that $[\mathrm{Gal}(N/\mathbb{R}) : H] = [N^H : \mathbb{R}]$ is odd -- in fact, $[N^H : \mathbb{R}] = 1$ by the first paragraph of this post, so $H = \mathrm{Gal}(N/\mathbb{R})$, so $[N : \mathbb{C}]$ is a power of $2$ since it divides $|H|$. If $N \neq \mathbb{C}$, then there is a degree-$2$ extension of $\mathbb{C}$, but no such extension can exist because every element of $\mathbb{C}$ has a square root. Therefore $N = \mathbb{C}$ as desired.
Now suppose there is an irreducible, monic polynomial $f(x) \in \mathbb{R}[x]$. As $\mathbb{C}$ is algebraically closed, $\mathbb{C}$ contains a root $\alpha$ of $f$, so $[\mathbb{R}(\alpha) : \mathbb{R}]  \leq [\mathbb{C} : \mathbb{R}] = 2$. Also $\mathbb{R}(\alpha) \cong \mathbb{R}[x]/(f)$ via a field isomorphism that is the identity on $\mathbb{R}$, so
$$ \deg f = [\mathbb{R}[x]/(f) : \mathbb{R}] = [\mathbb{R}(\alpha) : \mathbb{R}] \leq 2 $$
 A: You don't need algebraic closedness of $\mathbb{C}.$ All you need is that every polynomial has a complex root (this is proved in a particularly painful way here). If the root is real, you are cooking with gas, if it is complex, it's complex conjugate is also a root, so $(x-r)(r-\overline{r})$ is a degree two factor with real coefficients. Of course, since the first sentence also proves the fundamental theorem of algebra, I am not sure what the upside is. I suppose the real question is: can one prove the statement without using complex numbers at all. A couple of even more horrible ways of showing this were suggested in the answers to this: Proving (without using complex numbers) that a real polynomial has a quadratic factor
A: The question is: prove that if $f \in \mathbb{R}[x]$ has even degree $\geq 4$, then $f$ is reducible. Answering this question is essentially equivalent to proving the fundamental theorem of algebra. It's clear that the FTA implies this fact, and below I will show that this fact implies the FTA.

Suppose that the only irreducible polynomials in $\mathbb{R}[x]$ of even degree are quadratics. It follows from the intermediate value theorem (see Proof that every polynomial of odd degree has one real root for further discussion) that every irreducible polynomial in $\mathbb{R}[x]$ is either linear or quadratic.
Now consider $g \in \mathbb{C}[x] - \mathbb{C}$. We want to show $g$ has a root in $\mathbb{C}$. Take $F$ to be the splitting field of $g$ over $\mathbb{C}$, so that $F$ is a finite-degree extension of $\mathbb{C}$; it suffices to show that $F = \mathbb{C}$.
We know $F = \mathbb{R}(\gamma)$ by the primitive element theorem, so the minimal polynomial of $\gamma$ over $\mathbb{R}$ is an irreducible in $\mathbb{R}[x]$, so it must be linear or quadratic, so $[F : \mathbb{R}] \in \{1, 2\}$. Of course $F \neq \mathbb{R}$, so $2 = [F : \mathbb{R}] = [F : \mathbb{C}][\mathbb{C} : \mathbb{R}]$ implies $[F : \mathbb{C}] = 1$, so $F = \mathbb{C}$ as desired.
