We have a sentence about prime polynomials over a field $K$. It's stated that whenever a polynomial $f(x) \in K[x]$ with $deg f(x) \ge 2$ has a root $c$ in $K$, it's clear that $(x-c) \mid f(x)$ and thus it's no prime polynomial.
What I am not sure about is the opposite: If a polynomial doesn't have a root in K, is it automatically a prime polynomial? Is that true?
Thanks a lot for your help.
The answer is no: Consider the polynomial $(x^2+1)^2$ over $\mathbb R$. This is clearly no prime element but does not have any roots in $\mathbb R$. But the statement is true whenever the degree of your polynomial is at most 3, since in this case any factorization into prime factors will necessarily include a linear term (provided that the polynomial has not been prime to begin with).