Prime $p\in R$ remains prime in $R[x]$ (Gauss's Lemma)

Let $R$ be an integral domain , $p \in R$ be a prime element , then is $p$ also a prime element in $R[x]$ ?

I know it is true if $R$ is a UFD , because then the polynomial ring $R[x]$ is a UFD and showing $p$ is prime is equivalent to showing it is irreducible .

But what of $R$ is not a UFD ? Please help . Thanks in advance

Yes, use the fact that $R[x]/(p) = (R/p)[x]$. Then $R/p$ is an integral domain because $p$ is prime so the polynomial ring $(R/p)[x]$ is as well. And by definition $\mathfrak{p}$ an ideal of a ring $S$ is prime iff $S/\mathfrak{p}$ is an integral domain.

It's instructive to give a direct proof of $\,p\nmid A,B\,\Rightarrow\,p\nmid AB.\,$ Since $\,p\nmid A,B\,$ when reduced mod $p$ both have lead coefs $\,\color{#0a0}{a,b\not\equiv 0}\,$ so $AB$ has lead coef $\,\color{#c00}{ab\not\equiv 0}\,$ (by $p$ prime), hence $AB\not\equiv 0,\,$ i.e.

$\ \ \ \ \qquad{\rm mod}\ p\!: \ \ \ \begin{eqnarray} &&\ 0\ \not\equiv\ A\ \equiv\, \color{#0a0}a\, x^j\! + \:\cdots,\quad\ \ \:\! \color{#0a0}{a\not\equiv 0}\\ &&\ 0\ \not\equiv\ B\ \equiv\, \color{#0a0}b\, x^k\! + \:\cdots,\quad\ \ \, \color{#0a0}{b\not\equiv 0}\\ \Rightarrow\,\ &&0 \not\equiv AB \equiv \color{#c00}{ab}\ x^{j+k}\! + \:\cdots,\, \color{#c00}{ab\not\equiv 0}\end{eqnarray}$

i.e. primes $\,p\in R\,$ remain prime in $\,R[x]\,$ because the prime divisor property $\,\color{#0a0}{p\nmid a,b}\,\Rightarrow\ \color{#c00}{p\nmid ab}\,$ persists when multiplying leading coefficients. This is one form of Gauss's Lemma.

This is an elementwise view of the common proof sketched in Adam's answer (it's simply the proof of $\,D\,$ domain $\Rightarrow\,D[x]\,$ domain, in the special case that $\,D \cong R/p \cong R\bmod p,\,$ for $p$ prime).

Beware $ $ The proof depends crucially on $R$ being UFD. For if $R$ has an atom (irreducible) $\,p\,$ that is not prime then there exists $a,b\,$ such that $\,p\mid ab,\ p\nmid a,\ p\nmid b\,$ so $\,f= px+a,\,g = px+b\,$ are primitive but $\,p\mid fg\,$ so $\,fg\,$ is not primitive, so Gauss's Lemma fails. See here for more.