Theorem 23 10

Statement

Let $f(x) \in F[x]$, and let $f(x)$ be of degree $2$ or $3$. Then $f(x)$ is reducible over $F$ if and only if it has a zero in $F$.

Proof

If $f(x)$ is reducible so that $f(x)=g(x)h(x)$, where the degree of $g(x)$ and the degree of $h(x)$ are both less than the degree of $f(x)$, then since $f(x)$ is either quadratic or cubic, either $g(x)$ or $h(x)$ is of degree $1$. If, say, $g(x)$ is of degree $1$, then except for a possible factory in $F$, $g(x)$ is of the form $x-a$. Then $g(a)=0$, which implies that $f(a)=0$, so $f(x)$ has a zero in $F$.

Conversely, Corollary 23.3 shows that if $f(a)=0$ for $a \in F$, then $x-a$ is a factor of $f(x)$, so $f(x)$ is reducible.