Let $f$ be an irreducible (over $\mathbb{Q}$) polynomial in $\mathbb{Z}[x]$, $\deg (f)=3,4,5$. The Galois group of an irreducible polynomial $f\in \mathbb{Z}[x]$ acts transitively on distinct roots in $\mathbb{C}$ of $f$. Hence $Gal(f)\leq S_m$, and acts transitively on distinct $m$-roots, where $m\leq n$ for $n=\deg (f)$.
$m=3$, $S_3,A_3=\mathbb{Z}_3$
$m=4$, $S_4,A_4,D_4,\mathbb{Z}_4,K_4$ ($K_4$ is the Klein four group)
$m=5$, $S_5,A_5,D_5,\mathbb{Z}_5,Fr_5$ ($Fr_5$ is a Frobenius group)
For example, by Eisenstein criterion, the following polynomials are irreducible, but:
Determine over $\mathbb{Q}$, $Gal(x^3-3x+1)=\mathbb{Z}_3$, not $S_3$? Why?
Determine over $\mathbb{Q}$, $Gal(x^4+3x+3)=D_4$, not $A_4,K_4, S_4$? Why?
Also by determine coefficients, $x^4+8x+12$ is irreducible. Why $Gal(X^4+8x+12)=A_4$, over $\mathbb{Q}$, not $S_4,K_4, \mathbb{Z}_4$?
Is there any method to determine the Galois group?