> [!NOTE] Corollary > Let $\mathbb{C}[x]$ be the [[Ring of Polynomial Forms|ring of polynomials]] in $x$ over the [[Complex Numbers|complex numbers]]. Let $f\in \mathbb{C}[x].$ Then $f$ is [[Irreducible Polynomial|irreducible]] over $\mathbb{C}[x]$ iff $\deg(f)=1.$ **Proof**: Let $f\in \mathbb{C}[x].$ Suppose $\deg(f) \geq 1$ then by [[Fundamental Theorem of Algebra|FTA]], there exists $\alpha \in \mathbb{C}$ such that $f(\alpha) = 0$. By [[Polynomial Factor Theorem|factor theorem]], $f(x) = (x-\alpha)g(x)$ for some $g\in \mathbb{C}[x]$. Conversely, if $\deg(f)=1$ then it is irreducible over $\mathbb{C}[x]$: suppose $\deg(f)=1$ and $f=gh.$ Then by [[Degree of Product of Polynomials Over Integral Domain|degree of product of polynomials]], $1=\deg(g)+\deg(h)$ which implies that either the degree of $g$ or $h$ is zero.