Prove that if R is a UFD then R[x] is also a UFD.

Question

Prove that if $R$ is a UFD then $R[x]$ is also a UFD. I look at some proofs online but not able to understand much. Proof goes like this:

Let $R$ be a UFD and let $F$ denotes the field of fraction of $R$. The key to showing that $R[x]$ is a unique factorization domain is to compare factorizations in $R[x]$ with factorizations in the Euclidean domain $F[x]$. Call an element of $R[x]$ primitive if its coefficients are relatively prime. Any element $g(x) ∈ R[x]$ can be written as $$g(x) = dg_1(x),$$where $d ∈ R$ and $g(x)$ is primitive. Moreover, this decomposition is unique up to units of $R$. After this I am not able to understand and I am not able to understand the bold written text.

Reference : http://homepage.math.uiowa.edu/~goodman/22m121.dir/2005/section6.6.pdf

Answer 1

If $K$ is the field of fractions of $R$, then $K[x]$ is a UFD because it is Euclidean hence a PID.

So, every polynomial in $R[x]$ has a unique factorization in $K[x]$.

The crucial point is that this factorization is actually in $R[x]$.

The key result in this path is Gauss's lemma.