Let $F$ be a field, $f \in F[x]$ of degree 2 or 3. Prove that if $f$ has no roots, then $f$ is irreducible.
I am trying to use the contrapositive to prove this, as I was told it might be easier.
NTS: If $f$ has no roots, then $f$ is irreducible. Contrapositive: If $f$ is not irreducible, then $f$ has roots.
My attempt (converse of the contrapositive, I'm having trouble with proving the contrapositive. I can prove the converse of the contrapositive as shown below, but am stuck with proving the contrapositive. Any feedback would be appreciated! ):
pf. Let $F$ be a field, $f \in F[x]$ of degree 2 or 3. Then, $f$ has a root, say $a$, in $F$. Then by the Factor Theorem, $(x - a) \vert f$. So $f = (x-a)g$ for some $g \in F[x]$. Well, $g$ must have degree $\geq$ 1, so neither factor is a unit. Thus $f$ is not irreducible.