how should Cayley's resolvent be used in order to see if a given quintic equation is solvable or not? 
Hi. I chose a simple equation to work with: $x^5+x^4+x^3+x^2+x+1=0$ (which is solvable and $x=-1$).
the amounts of $p$, $q$, $r$ and $s$ where $\frac{3}{5}$, $\frac{14}{25}$, $\frac{87}{125}$ and $\frac{2604}{3125}$ respectively.
I calculated $P$ to be $P=z^3-15z^2-69z+1235$ and $\Delta$ to be $\Delta=1296$.
as the picture says, $P^2-1024z\Delta$ (Cayley's resolvent) should have rational root(s) in $z$ in order to the first qintic function be solvable.
but the problem is that $P^2-1024z\Delta=0$ is $z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=0$ which is even worse and impossible to solve!
what should I do now?!

Hint:
We have the following factorization:
$$
z^6-30z^5+87z^4+4540z^3-32289z^2-1497534z+1525225=
(z^2 + 22z + 169)(z^2 - 26z + 361)(z - 1)(z - 25).
$$
Of course, the splitting field of 
$$
x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}
$$
is $\Bbb Q(\zeta_6)$, so that the Galois group is  solvable.
