TITLE: Irreducible polynomial means no roots?
QUESTION [18 upvotes]: If a polynomial is irreducible in $R[x]$, where $R$ is a ring, it means that it does not have a root in $R$, right?
For example, to say that a polynomial $f(x)\in\mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ is equivalent to say that $f(x)$ does not have any rational root. I just want to make sure.

REPLY [13 votes]: I know there's an accepted answer here, but I just wanted to add in something to clarify a couple answers for newcomers:
If $F$ is a field, $f(x)\in F[x]$ is reducible if and only if $f(x)$ has a zero in $F$, but this is only always true for polynomials of degree 2 and 3. 
Mark Bennet gives a decent counterexample to the generalized claim, and note that the polynomial he uses is degree 4.
However, things are a bit different when you're working in $Z_n$ ($Z/nZ$). You can check for reducibility by testing if $f(n)=0$ for $n\in[0,n-1]$.
For example, $f(x)=x^3+1\in Z_9[x]$ is reducible over $Z_9$ because $f(2)=0$.