TITLE: Is $f(x)$ reducible if $f(a)=0$
QUESTION [6 upvotes]: I am confused about this seemingly trivial question: 

If $f(a) = 0$ for some $a\in D$, then when is $f(x)$ reducible in $D[x]$? ($D$ is an integral domain).

My answer: Always.
Let $f(a)=0$. Then $f(x)$ has $(x-a)$ as factor. So $f(x)$ can be written as $f(x) = (x-a)h(x)$. Now what is the guarantee that $h(x)\in D[x]$? I believe it is the division algorithm for polynomials (remainder = 0). Am I correct in proving $h(x)\in D[x]$?

REPLY [4 votes]: "Always" is wrong. For example $f(x)=x-a$ is irreducible.