Exploring 3-cycle points for quadratic iterations How do you factorise $z^8 +4cz^ 6 + (6c^2 +2c)z^4 + (4c^3 +4c^2 )z^2 - z+ (c^4 +2c^3 +c^2 +c)$?
I want to find the 3-cycle points for the quadratic iteration $z \rightarrow z^2 + c$. In order to find it I have to solve the above polynomial. I have factorised into two factors 
$(z^2 - z + c) (z^6+z^5+(3c+1)z^4+(2c+1)z^3+(3c^2+3c+1)z^4+((c+1)^2)z +(c^3+2c^2+c+1).$.
Note: 
I did missed something in the factors previously. Now it is corrected.
 A: The Galois group of such a polynomial of degree $6$ will not be $S_6$.
As soon as you know one of the roots $z$, you know two of the others, by applying $z \mapsto z^2+c$ twice.
The splitting field of $P$ should then generally be of degree $6 \times 3 = 18$.
Let $\zeta$ be a primitive thrid root of unity.
If you call the roots $z_{1,1},z_{1,\zeta},z_{1,\zeta^2},z_{2,1},z_{2,\zeta},z_{2,\zeta^2}$ where $z_{i,j}^2 + c = z_{i,\zeta j}$, any automorphism sending $z_{i,j}$ to $z_{i',j'}$ has to send $z_{i,\zeta j}$ to $z_{i',\zeta j'}$, so to determine an automorphism completely you only need to know the images of $z_{1,1}$ (among $6$ possibilities) and $z_{2,1}$ (among $3$ possibilities).
Let $H$ be the subgroup that does not switch the two $3$-cycles. $H$ is normal of index $2$ in the Galois group $G$, and $H$ is in fact isomorphic to $(\Bbb Z/3\Bbb Z)^2$.
So $G$ is a solvable group.
Let $P_1(z) = (z - z_{1,1})(z - z_{1,\zeta})(z - z_{1,\zeta^2})$. $P_1$ is invariant by $H$, so its coefficients should all belong in the fixed field of $H$, a quadratic extension $K$ of $\Bbb Q(c)$
First we should figure out what is $K$.
The extension ramifies when we have $z_{1,j} = z_{2,j}$, so a necessary condition is that the polynomial has multiple roots. 
We can compute the discriminant of $P$. Amazingly, it's simpler than we should expect, as it factors as $\Delta = (4c+7)^3(16c^2+4c+7)^2$.
There are two ways that $P$ can have multiple roots. Either the two $3$-cycles are equal (then we should obtain a multiplicity $3$ root of $\Delta$), either one $3$-cycle collapses into a fixpoint of $z \mapsto z^2+c$ (then we should obtain a multiplicity $2$ root of $\Delta$). 
We can confirm that the degree $2$ factors belong in the second case by computing the resultant of $z^2-z+c$ and $P$ (which gives $16c^2+4c+7$). So the only time when the quadratic extension can ramify is when $4c+7 = 0$, hence the extension is $\Bbb Q(c)(\sqrt{q(4c+7)})$ for some squarefree integer $q$.
When we plug $c=0$, $P_0(z)= z^6+z^5+z^4+z^3+z^2+z+1$, of which we know the roots.
The two degree $3$ factors over $K$ must then be $(X - \zeta_7)(X - \zeta_7^2)(X-\zeta_7^4)$ and $(X - \zeta_7^3)(X - \zeta_7^5)(X-\zeta_7^6)$.
Since the only quadratic extension contained in the cyclotomic extension $\Bbb Q \subset \Bbb Q(\zeta_7)$ is $\Bbb Q(\sqrt{-7})$, we must have $K = \Bbb Q(c)(\sqrt{-4c-7})$.
Next we have to find out the factorisation of $P$ over $K$.
Using Gauss' lemma, the coefficients of the factors have to be polynomials in $c$ (to be precise, they have to be in $\Bbb Z[c, \frac{1+\sqrt{-4c-7}}2]$)
By making simple assumptions on the degree of the coefficients, and using the factorisation at $c=0$ ($P_0(X) = (X^3 + \frac{1+\sqrt{-7}}2X^2 + \frac{-1+\sqrt{-7}}2X - 1) 
 (X^3 + \frac{1-\sqrt{-7}}2X^2 + \frac{-1-\sqrt{-7}}2X - 1)$)
we can quickly find
$$P(X) = \left(X^3 + \frac{1+\sqrt{-4c-7}}2X^2 + \frac{2c-1+\sqrt{-4c-7}}2X + \frac{-c-2+c\sqrt{-4c-7}}2\right) \\
 \left(X^3 + \frac{1-\sqrt{-4 c-7}}2X^2 + \frac{2c-1-\sqrt{-4c-7}}2X + \frac{-c-2-c\sqrt{-4c-7}}2\right)
$$
Finally we have to solve each cubic polynomial (which has cyclic Galois group)
To do that, let $w_i = (z_{i,1} + \zeta z_{i,\zeta} + \zeta^2 z_{i,\zeta^2})$. The conjugates of $w_i$ are $\zeta w_i$ and $\zeta^2 w_i$, which means that $w_i^3 \in K(\zeta)$. The same happens with $v_i = (z_{i,1} + \zeta^2 z_{i,\zeta} + \zeta z_{i,\zeta^2})$ and in fact $v_i^3$ is the image of $w_i^3$ after switching $\zeta$ with $\zeta^2$. By summing two appropriate cube roots of those two, you obtain $2z_{1,1} - z_{1,\zeta} - z_{1,\zeta^2}$. Add to it the sum of the three roots (which is a coefficent of $P_1$, in $K$) and divide by $3$ to get $z_{1,1}$.
