Reducibility test for degrees 2 and 3

$F$ : field

Reducibility test for degrees 2 and 3 is as follows,

$$f(x) \in F[x] \text{ and } deg\ f(x) = 2\text{ or }3$$ $$\implies$$ $$f(x)\text{ is reducible over }F\text{ iff }f(x)\text{ has a zero in }F$$

Suppose that $f(x) = g(x)h(x)$, where $g(x),\ h(x) \in F[x]$. Since that, polynomials of zero degree are the only unit over integral domain, forms containing zero polynomial are not our concern. Hence, $1 \leq g(x),\  h(x) < deg\ f(x)$

Since, in integral domain, deg $f(x) = deg\ g(x) + deg\ h(x)$ and $deg\ f(x)$ is $2$ or $3$, at least one of g(x) and h(x) has degree 1. Say $g(x) = ax + b$. Then, clearly, $-a^{-1}b$ is a zero of $g(x)$ and therefore a zero of $f(x)$.

Conversely, suppose that $f(a) = 0, a \in F$. Then by the Factor Theorem, we know that $x - a$ is a factor of $f(x)$ and, therefore $f(x)$ is reducible over $F$.