Find Galois Group of $4x^4+5x^3-9$ over $\mathbb{Q}$ I proceeded as follow: 
first of all i noticed that $4x^4+5x^3-9$=$(x-1)*(4x^3+9x^2+9x+9)$. So the polynomial has the same galois group of $(4x^3+9x^2+9x+9)$ which is a cubic so it should be either $A_3$ or $S_3$. (Am i wrong? why it has to be one of this two?).
Then i evaluated the discriminant that seems to be negative, and i think the group is $S_3$, but i'm doing this without having truly understood the reason..
 A: Suppose $f(x) \in \mathbb{Q}[x]$ is a monic irreducible polynomial of degree $d$, and let $\alpha_1, \alpha_2, \cdots, \alpha_d$ denote its roots in $\overline{\mathbb{Q}}$. Consider the product $$P := \prod_{i < j} (\alpha_i - \alpha_j)$$ Note that $P^2$ is the discriminant of $f$.
If an element $\sigma \in \text{Gal}(f) \subset S_{d}$ has $\sigma \in A_d$, then $\sigma(P) = P$, since $\sigma$ can be decomposed into an even number of transpositions. Thus, if $\text{Gal}(f) \subset A_d$, then $\sigma(P) = P$ for all $\sigma \in \text{Gal}(f)$, hence $P \in \mathbb{Q}$. In particular, the discriminant $P^2$ must be a rational square.
Lemma: If $\text{Gal}(f) \subset A_d$, then the discriminant of $f$ must be a rational square.
Now, let $f(x) = 4x^3 +9x^2 + 9x + 9$. To compute the discriminant, it suffices to compute the discriminant of $f(x+a)$ for any $a\in \mathbb{Q}$ (translating roots does not change the discriminant).
Note that $f(x+3/4) = 4x^3 + (9/4)x + 45/8$, i.e. by translating we have "dampened" the $x^2$ term. Now, we compute the discriminant of $f/4$ to be $-13851/256$, which is certainly not a rational square.
Thus, $\text{Gal}(f) = S_3$.
