If R[x] is a UFD, then R is a UFD. I see the statement that if R is UFD then R[x] is a UFD, but is the converse of the statement, which is that if R[x] is a UFD, then R is a UFD, true?
 A: I have written a full proof at the bottom (in case the hint isn't enough). 
Hint: Since $R\lbrack x\rbrack$ is a UFD, it is (in particular) an integral domain. Hence degree considerations work -- i.e. we always have equality $\deg (fg)=\deg f+\deg g$, unlike in the general case where 
$\deg (fg) \leq \deg f + \deg g$. 
It follows that whenever we factorize elements of $R$ in $R\lbrack x\rbrack$, we actually get (by degree considerations) a factorization in $R$.
Also, elements of $R$ which are irreducible in $R\lbrack x\rbrack$ must in particular be irreducible in $R$. This is a fairly straightforward argument, just remember that the units of $R\lbrack x\rbrack$ are simply the units of $R$ viewed as constant polynomials.
Addendum. In general, a subring of a UFD need not be a UFD. As mentioned in an answer to this question, the easiest counterexamples are constructed by choosing an integral domain which is not a UFD, then noting that this ring is a subring of its field of fractions (which is trivially a UFD). This is why we need arguments like the one sketched above, which in one way or another use the fact that $R\lbrack x\rbrack$ consists of polynomials over $R$.
So that you don't have to take my word for it: There are, of course, many examples of non-UFD integral domains. Trotter observed (in an article cited by Lang) that the well-known identity $\cos^2x + \sin^2x = 1$ implies the non-unique irreducible factorizations 
$$\sin^2(x)=(1+\cos x)(1-\cos x)$$
 in the ring of trigonometric polynomials $\mathbb R \lbrack \sin x,\cos x\rbrack$. Meanwhile, $\mathbb R \lbrack \sin x,\cos x\rbrack$ is easily seen to be an integral domain.
Here is the promised proof that if $R\lbrack x\rbrack$ is a UFD, then $R$ is a UFD.
Proof. This (simple) proof is divided into three steps. Recall that the units of $R\lbrack x\rbrack$ are precisely the units of $R$. Also note that since $R\lbrack x\rbrack$ is an integral domain and $R\subset R\lbrack x\rbrack$, then $R$ is an integral domain. Therefore, we just need to show existence and uniqueness of irreducible factorizations in $R$. 
Step 1) We claim that $p\in R$ is $R$-irreducible (i.e. irreducible as an element of $R$) iff $p$ is $R\lbrack x\rbrack$ irreducible. We may assume $p\neq 0$. The "if" part is trivial. Indeed, if $p = ab$ for $a,b\in R$, then $p = ab$ is also a factorization in $R\lbrack x\rbrack$. Hence wlog $a$ must be a unit in $R\lbrack x\rbrack$, so also a unit in $R$. As for "only if", suppose $p$ is $R$-irreducible and $p = fg$ is a factorization in $R\lbrack x\rbrack$. Then
\begin{align*}
\deg f + \deg g = \deg p = 0  &\Rightarrow \deg f = \deg g = 0 \\
&\Rightarrow f,g\in R.
\end{align*}
Hence $p = fg$ is a factorization in $R$, so wlog $f$ is a unit in $R$, thus also a unit in $R\lbrack x\rbrack$. 
Step 2) If $a\in R$, then $a$ has a factorization
\begin{align}\label{factors}
a = p_1\dots p_n \tag{1}
\end{align}
where $p_i$ is an irreducible element of $R\lbrack x\rbrack$ for $i = 1,\dots ,n$. As in step 1,
$$
\deg p_1 + \dots + \deg p_n = \deg a = 0 \Rightarrow p_i \in R\quad i = 1,\dots ,n.
$$
By step 1, each $p_i$ is therefore $R$-irreducible. Then \eqref{factors} is an irreducible factorization in $R$. 
Step 3) It remains to show uniqueness. Suppose $a\in R$ and $p_1,\dots ,p_n$ and $q_1,\dots ,q_m$ are irreducible elements in $R$ with
\begin{align}\label{twofactors}
p_1\dots p_n = a = q_1\dots q_m.\tag{2}
\end{align}
By step 1), \eqref{twofactors} gives two $R\lbrack x\rbrack$-irreducible factorizations of $a$. But $R\lbrack x\rbrack$ is a UFD, so $m = n$ and there is a permuation $\sigma \colon \lbrace 1,\dots ,n\rbrace\rightarrow \lbrace 1,\dots ,n\rbrace$ and units $u_i$ ($i=1,\dots ,n$) in $R\lbrack x\rbrack$ so that
\begin{align}\label{perm}
p_i = u_iq_{\sigma (i)}\quad i=1,\dots ,n. \tag{3}
\end{align}
But $u_i$ is also a unit in $R$, so \eqref{perm} shows that $p_1\dots p_n$ and $q_1\dots q_m$ are the same factorization (up to permutation and unit) in $R$. This completes the proof. QED.
A: Hint: Suppose otherwise; i.e. there exist an element of $R$ which has distinct prime factorisations ("really" distinct, they're not reorderings/unit multiplications of each other). Then this element is also an element of $R[x]$. All we need to show is that primes in $R$ remain primes when viewed as an element of $R[x]$, and that non-units remain non-units, in order to contradict that $R[x]$ is a UFD. 
