# Prove that if $f$ has no roots, then $f$ is irreducible.

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.

• Prove that if $f$ is reducible, then it can be divided by a polynomial of degree $1$. – Suzet Dec 1 '19 at 3:32
• 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. – Captain Lama Dec 1 '19 at 3:55

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.