$x^4+8x+12$ has Galois group $A_4$ over $\mathbb{Q}$

No integer divisor of $12$ is a root, so there is no rational linear factor. Modulo $5$, one has $x^4+8x+12=(x-4)(x^3+4x^2+x+2)$; the cubic has no root in ${\mathbb{F}}_5$ and is irreducible. A monic factorization into two rational quadratics would reduce to a quadratic-by-quadratic factorization modulo $5$, contradicting the displayed irreducible factorization. Gauss's lemma therefore makes the quartic irreducible over $\mathbb{Q}$.

The resolvent is $y^3-48y-64$. Modulo $5$ it is $y^3+2y+1$, whose values at all elements of ${\mathbb{F}}_5$ are nonzero; hence the cubic resolvent is irreducible over $\mathbb{Q}$.

Its discriminant, and hence the quartic discriminant, is $-4(-48{)}^3-27(-64{)}^2=331776={576}^2$, which is nonzero.

Steps 1.1, 1.2, and 1.3 satisfy [L1], so the quartic has Galois group $A_4$.