Is there alternative factoring of a quintic equation? In a paper here
the author appears to be able to factor a Bring-Jerrard quintic making
$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$
become
$$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$
but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:
$$n^5-m^4n+\frac{P}{2m}=0$$
given that I will know the values of $P$ and $m$.
Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?

It seems that the claim of the paper is not true. 
The author finally got the following equation
$$b^4-2b^3\bigg(\frac{q-p+\sqrt{q^2-q}}{q-\sqrt{q^2-q}}\bigg)+q-p+\sqrt{q^2-q}=0$$
Here, let us consider one example. We have
$$x^5-31x+30=(x^3+3x^2+7x+15)(x^2-3x+2)$$
This means that one of the possible values of $b$ is $3$ for $(p,q)=(-31,30)$.
However, the above equation does not have a solution $b=3$ for $(p,q)=(-31,30)$.

Since we have
$$(x^3+bx^2+cx+d)(x^2+ex+f)$$
$$=x^5+(b+e)x^4+(eb+c+f)x^3+(bf+ec+d)x^2+(cf+ed)x+df=0$$
if we compare this with $x^5+px+q$, then we get the following system
$$\begin{cases}b+e=0
\\eb+c+f=0
\\bf+ec+d=0
\\cf+ed=p
\\df=q\end{cases}$$
from which we want to represent $b,c,d,e,f$ by $p,q$.
Now, we have
$$\begin{align}&\begin{cases}b+e=0
\\eb+c+f=0
\\bf+ec+d=0
\\cf+ed=p
\\df=q\end{cases}\\\\&\stackrel{\text{eliminating $e$}}{\implies} 
\begin{cases}e=-b
\\(-b)b+c+f=0
\\bf+(-b)c+d=0
\\cf+(-b)d=p
\\df=q\end{cases}
\\\\&\stackrel{\text{eliminating $f$}}{\implies}\begin{cases}e=-b
\\df=q
\\-b^2d+cd+q=0
\\bq-bcd+d^2=0
\\bd^2=cq-pd
\end{cases}
\\\\&\stackrel{\text{eliminating $b$}}{\implies}
\begin{cases}e=-b
\\df=q
\\bd^2=cq-pd
\\c^2dq^2-cqd^2p+d^3p^2-d^2pcq-cd^5-qd^4=0
\\c^2dq^2-cq^3+dpq^2-cd^2pq-d^4q=0
\end{cases}
\\\\&\stackrel{\text{eliminating $c$}}{\implies}
\begin{cases}e=-b
\\df=q
\\bd^2=cq-pd
\\c(-d^2pq-d^5+q^3)=dpq^2-d^3p^2
\\d^{10} + p qd^7 +  p^3d^6 - 2  q^3d^5 -  p^2 q^2 d^4- p q^4d^2 + q^6=0
\end{cases}\end{align}$$
So, we have to solve the following equation for $d$ : 
$$d^{10} + p qd^7 +  p^3d^6 - 2  q^3d^5 -  p^2 q^2 d^4- p q^4d^2 + q^6=0$$
whose degree is $10$.
In conclusion, if we want to find $b,c,d,e,f$ such that
$$x^5+px+q=(x^3+bx^2+cx+d)(x^2+ex+f)$$
then, in general, we have to solve an equation whose degree is $10$.
