If a polynomial has a rational root is it automatically reducible? This seems obvious, but I just can't crack it.
Let K be a field and $F(X) \in K[X]$ be a polynomial. Does $F(a)=0$ for some $a\in K$ imply that F(X) is reducible. Clearly, by the fundamental theorem of algebra, $F(X) = (X-a)G(X)$ for some $G(X) \in \bar{K}[X]$, but does it follow that actually $G(X) \in K[X]$?
 A: Yes: if $a\in K$ then $G \in K[X]$ because the Euclidean division algorithm of $F$ by $X-a$ works entirely over $K$.
Indeed, you get $F(X)=(X-a)Q(X)$ with $Q\in K[X]$. But $(X-a)G(X)=F(X)=(X-a)Q(X)$ over $\bar K$ and you can cancel $X-a$ to conclude $G=Q \in K[X]$.
The fundamental theorem of algebra plays no role here. The Euclidean division algorithm works over any field (and over any commutative ring when the divisor is monic).
A: Note $\ $ There seems to be varying interpretations of the question. I read it as asking that if we somehow find a factorization $\rm\:f=(x−a)\:g\:$  for $\rm\:g\in \hat K[x]\:$ over some extension of the coefficient ring $\rm\:\hat K\supset K,\:$ then does this necessarily yield a factorization over $\rm K,\:$ i.e. is $\rm\:g\in K[x]$? Some readers interpreted the question more simply as: must there exist $\rm\:g\in K[x],\:$ vs. $\rm\:g\in \hat K[x]\: \Rightarrow\: g\in K[x].$
Hint $\ $ It follows from the uniqueness of the quotient (and remainder) in the division algorithm (which is the same in $\rm\:K[x]\:$ and $\rm\:\bar K[x],\:$ using the polynomial degree as the Euclidean valuation). Thus since dividing $\rm\:f\:$ by $\rm\:x-a\:$ in $\rm\:\bar K[x]\:$ leaves remainder $0$, by uniqueness, the remainder must also be $0$ in $\rm K[x]\:,\:$ i.e. $\rm\:x-a\ |\ f\ $ in $\rm \bar K[x]\:$ $\:\Rightarrow\:$ $\rm\:x-a\ |\ f\ $ in $\rm K[x]\:.\:$  
This is but one of many examples of the power of uniqueness theorems for proving equalities.
