Deducing that $\mathbb Z[2 \sqrt{2}]$ is not a $UFD.$

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Here is the question I am trying to understand the solution of the last part in it:

Let $R$ be an integral domain with quotient field $F$ and let $p(x)$ be a monic polynomial in $R[x].$ Assume that $p(x) = a(x) b(x)$ where $a(x)$ and $b(x)$ are monic polynomials in $F[x]$ of smaller degree than $p(x).$ Prove that if $a(x) \notin R[x]$ then $R$ is not a $UFD$. Deduce that $\mathbb Z[2 \sqrt{2}]$ is not a $UFD$.

Here is a solution I found here:

$\mathbb{Z}[2\sqrt2]$ is not a UFD.

My question is:

What is the quotient field of $\mathbb Z[2 \sqrt{2}][x]$? Why the given factors in the solution are in the quotient field of $\mathbb Z[2 \sqrt{2}][x]$?

Can someone clarify this to me please?