Let $R$ be an integral domain and $f\in R[x]$ be a monic polynomial of degree $2$ or $3$. Let $K$ be the field of fractions of $R$. Are the following statements true?
$f$ is irreducible in $K[x]$ $\iff$ $f$ has no root in $K$
$\Downarrow$
$f$ is irreducible in $R[x]$ $\iff$ $f$ has no root in $R$
If $R$ is integrally closed (not necessarily a UFD), we already have
$f$ is irreducible in $K[x]$ $\iff$ $f$ has no root in $K$
$\Updownarrow$
$f$ is irreducible in $R[x]$ $\iff$ $f$ has no root in $R$
For example, $x^2+1 \in \mathbb Z[2i][x]$ has the roots $\pm i$, which are not in $\mathbb Z[2i]$ but in its field of fractions $\mathbb Q(i)$. In other words, $\mathbb Z[2i]$ is not integrally closed.
$$\begin{matrix}[1]&\Leftrightarrow&[2]\\ \Downarrow&&\\ [3]&\Rightarrow&[4]\end{matrix}$$ holds indiscriminately for all domains $R$ and polynomials $f$: for $[3]\Rightarrow [4]$, you can prove by induction on $\deg f$ that if $g\in R[x]$ is monic and $f\in R[x]$ is such that $\frac fg\in K[x]$, then $\frac fg\in R[x]$. If $a\in R$ is a root of $f\in R[x]$, then $x-a$ divides $f$ in $K[x]$; therefore, $\frac{f}{x-a}\in R[x]$.
In the special case of $f$ monic with $\deg f\in \{2,3\}$, we also have $[4]\Rightarrow [3]$, because if $f$ is reducible in $R[x]$, then its non-trivial divisor of least degree can only have degree $0$ or $1$. Since $f$ is monic and $R$ is a domain, all its divisors must have invertible leading terms: i.e., we can assume that its divisor of least degree is monic. Therefore, its divisor of least degree is monic of degree $1$. I.e., there is some $a\in R$ such that $f=(x-a)g$.