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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.

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  • $\begingroup$ Prove that if $f$ is reducible, then it can be divided by a polynomial of degree $1$. $\endgroup$ – Suzet Dec 1 '19 at 3:32
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    $\begingroup$ Your proof is not correct: what you prove is that a polynomial of degree at least $2$ with a root is reducible. It is not the contrapositive but rather the converse of the statement you want to prove. $\endgroup$ – Captain Lama Dec 1 '19 at 3:55
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Suppose that $f$ is reducible, $f=gh$ where $deg(f)+deg(g)=3$ and $deg(f),deg(g)>0$ this implies that $deg(f)=1$ or $deg(g)=1$ and $f$ has a root.

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