<s>[INST] 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 $p,q$?
 [/INST] The usual way to solve that would be to expand 
$(x^3 +bx^2 +cx+ d)(x^2 + ex + f)=x^5 + (e + b)x^4 + (eb+c+f)x^3+(bf + d + ce)x^2 + (ed+cf)x + fd $
Want this to equal to $x^5 + px +q$ for all $x$ which means the two ploynomails are equal and thats only true if the coefficients for the same powers are equal so 
$$\begin{array}{cccCC}
e+b &=&0 \Rightarrow &e&=&-b \\
eb+c+f &=& 0\Rightarrow &c+f&=&b^2\\
bf+d+ce&=&0 \Rightarrow &d+b(f-c)&=&0 \\
ed+cf &=&p\Rightarrow &-bd+cf&=&p\\
fd &=& q\Rightarrow &f &= &\frac{q}{d}
 \end{array}$$
$$\begin{array}{ccc} 
cd + q &=& db^2 \\
d^2 + bq-bdc&=&0 \\
-bd^2+cq&=& dp\end{array}$$
Lets eliminate $c$ 
$$\begin{array}{ccc} 
cbdq + q^2b & = & qdb^3 \\ 
-bdcq + qd^2+bq^2 & = & 0 \\ 
cqbd - b^2d^3 & = &  d^2bp \\
\end{array}$$
Then 
$$\begin{array}{ccc}
q^2b+qd^2+bq^2 &=& qdb^3\\ qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$
We need to solve the two equations 
$$\begin{array}{ccc} 
2qb + d^2 &=& db^3 \\ 
qd^2+bq^2-b^2d^3&=&d^2bp \end{array}$$
</s>