Galois groups of quintic polynomials 
These two exercises are from Basic Algebra I; I'm not asking for the solution of one of them, I just want to know in general how $P (x) $ is involved in finding $G_f $ (that should be the difficult part of the exercise). I thought a bit but I didn't come up with any idea, so  I would like a hint. Thanks
 A: The polynomial $$\theta(y) = \prod_{i \le i \le j \le 5}(y - (r_i + r_j))$$ is a resolvent.
It is using the expression $r_1 + r_2$. A resolvent can be made using other expressions like $(r_1 - r_2)(r_1 - r_3)(r_2 - r_3)$ or $r_1 r_2 + r_3 r_4$.
A group that acts by permuting the roots $\{r_i\}$ can also act on expressions of the roots, for example if $\sigma r_1 = r_4$ and $\sigma r_2 = r_3$ then $\sigma (r_1 + r_2) = r_4 + r_3$.
The polynomial $\theta$ is constructed by multiplying together all the different values of $y - (r_1 + r_2)$ that we get by letting a permutation group act on it.
The resulting expression will be unaffected by permutations since it has been completely symmetrized. In Galois theory this means that it will exist in the base field i.e. it will have rational coefficients.
But the expression that we started with already has a small amount of its own symmetry. Let $H \le S_n$ be the group of permutations that leaves the expression fixed. Then symmetrizing it will produce $r = |S_n : H|$ terms, and give you a degree $r$ resolvent.
The resolvent will have a rational roots and factor completely over the base field iff the expressions (each of the permutations of $r_1 + r_2$) lie in the base field.

A simple example to work out is the discriminant for the cubic which is constructed from the expression $(r_1 - r_2)(r_1 - r_3)(r_2 - r_3)$. This expression is invariant under all even permutations $A_3$ but changes sign on odd permutations. The resultant you get for this is
$\theta(y) = (y - (r_1 - r_2)(r_1 - r_3)(r_2 - r_3))(y + (r_1 - r_2)(r_1 - r_3)(r_2 - r_3)) = y^2 - \Delta$
And this is why Galois group of a polynomial will be contained in $A_n$ if its discriminant is a square.

In problem 7 we have a square discriminant so the Galois group lies inside $A_5$, it must be one of $C_5 \le D_5 \le A_5$. It has been shown that the resolvent splits into two irreducible factors. This tells us something about the Galois group.
What this tells us is that the 10 conjugates of $r_1 + r_2$ are partitioned into two orbits by the Galois group. This could not happen with $A_5$ since it's Galois group acts transitively on our expressions.
What I am a little confused by is that we can get the same splitting behavior from the same resolvent for a quintic with $C_5$ Galois group. Based on this I would say that we cannot distinguish between $C_5$ and $D_5$ Galois groups using this resolvent.
The given polynomial does have 1 real root and thus 2 pairs of conjugate roots implying that there is an element of order 2 in the Galois group. Or we could use Dedekinds lemma on cycle type decomposition with $x^5 - 5x + 12 \equiv x (x+1)^4 \pmod 2$
The action of $D_5$ on terms $r_i + r_j$ can be visualized as an action on  lines of the pentagon. edges stay edges and diagonals stay diagonals so this action is intransitive.
