I know that if $R$ is a UFD and $\mathfrak{p} \subset R$ a prime ideal and $f$ a polynomial in $R[X]$ with content 1, s.t. $\deg(f)=\deg(\bar{f})$ then:
$\bar{f}$ is irreducible in $R/\mathfrak{p}[X] \Rightarrow$ f is irreducible in $R[X]$
(where $\bar{f}$ is $f$‘s image under the canonical homomorphism $R[X] \rightarrow R/\mathfrak{p}[X] $).
I‘ve now come upon the following example:
With $f=X^5+4X^2+14X+40 \in \mathbb{Z}[X]$ we get:
$\bar{f} = (X^3+2X+1)(X^2+1)$ in $\mathbb{Z}/(3)[X] $ and $\bar{f} = (X^4+4X+4)X$ in $\mathbb{Z}/(5)[X] $
Now since one is a decomposition into polynomials of degree 3 and 2 and the other one into polynomials of degree 4 and 1 $f$ must be irreducible in $\mathbb{Z}[X]$.
I dont quite understand why this works. I get that $f=gh$ in $R[X]$ induces a decomposition $\bar{f}=\bar{g}\bar{h}$ in $R/\mathfrak{p}[X]$ with $\deg(g)=\deg(\bar{g}),\deg(h)=\deg(\bar{h})$. So if the decomposition in $R/\mathfrak{p}[X]$ were unique an existing decomposition like $\bar{f}=\bar{g}\bar{h}$ with $\deg(\bar{h})=3, \deg(\bar{g})=2$ would imply that $f$ is either irreducible in $R[X]$ or can only be decomposed as a product of a polynomial of degree 3 and one of degree 2. But while $R[X]$ is a UFD, how do we know that $R/\mathfrak{p}[X]$, or rather $R/\mathfrak{p}$ is one (which would have to be the case s.t. such a decomposition would have to be unique).
A quick search shows me that whether $R/\mathfrak{p}$ is a UFD is a rather complicated question, so I must be missing some simpler argument. Any help is appreciated:)