> [!NOTE] Corollary > Let $\mathbb{C}[x]$ be the [[Ring of Polynomial Forms|ring of polynomials]] in $x$ over the [[Real numbers|real numbers]]. Let $f\in \mathbb{C}[x].$ Then $f$ is [[Irreducible Polynomial|irreducible]] over $\mathbb{R}[x]$ iff $\deg(f)=1$ or $\deg(f)=2$ with $f=ax^{2}+bx+c$ for some $a,b,c\in \mathbb{R}$ where $b^{2}-4ac<0.$ ***Proof***: It's clear that polynomials in $\mathbb{R}[x]$ which either have degree $1$ or have the form $ax^{2}+bx+c$ with $b^2-4ac<0$ are irreducible over $\mathbb{R}.$ Conversely, if $f\in \mathbb{R}[x]$ has the form $ax^{2}+bx+c$ with $b^{2}-4ac \geq 0$ then it is in fact reducible over $\mathbb{R}$ because it has real roots. Also if $\deg(f)\geq 2$ then by [[Fundamental Theorem of Algebra|FTA]], there exists $\alpha\in \mathbb{C}$ such that $f(\alpha)=0$ which implies $f(\bar{\alpha})=0.$ By repeated application of the [[Polynomial Factor Theorem|factor theorem]], this means that $f(x)=(x-\alpha)(x-\bar{\alpha})g(x)$ for some $g\in \mathbb{C}[x]$ with degree greater than zero (by [[Degree of Product of Polynomials|degree of product]]). But $(x-\alpha)(x-\bar{\alpha})\in \mathbb{R}[x].$ Thus by [[Division with remainder for integers|division with remainder]], there exists $q,r\in \mathbb{R}[x]$ with $\deg(r)<2$ such that $f(x)=(x-\alpha)(x-\bar{\alpha})q(x)+r(x).$This means that $(x-\alpha)(x-\bar{\alpha})(g(x)-q(x))=r(x)$If $g(x)-q(x)\neq 0$ then the degree of the LHS is at least $2$ and that of the RHS is less than $2,$ a contradiction. It follows that $g=q\in \mathbb{R}{[x]}$ and so $f$ is reducible over $\mathbb{R}.$