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For $P(x)=x^3+4x-7$, the derivative $P^{\prime }(x)=3x^2+4$ is positive everywhere, so the mean value theorem makes $P$ strictly increasing. Since a polynomial is a continuous function, the intermediate value theorem and the values $P(1)=-2$, $P(2)=9$ locate its unique real root in $(1,2)$. By the rational root theorem, a rational root of a monic integer polynomial must be an integer. There is no integer in $(1,2)$, so the unique real root is an irrational number. The denominator part of that theorem also follows directly: if $a/b$ in lowest terms were a root, $a^3+4a{b}^2-7b^3=0$ would force $b\mid a^3$, hence $b=1$.

Let $G$ be the Galois group of a polynomial of the irreducible quartic. Irreducibility makes $G$ a transitive subgroup of $S_4$. The three roots of the given cubic resolvent of a quartic correspond to the three partitions of the four quartic roots into two unordered pairs, and the kernel of the $S_4$-action on these partitions is the Klein four-group $V_4$. If the resolvent has Galois group $S_3$, the image of $G$ in this action is all of $S_3$. Among the transitive subgroups of the symmetric group on four points, only $S_4$ has this image. Hence

Now put $f_p(t)=t^4+pt+p$. The Eisenstein criterion at $p$ proves that $f_p$ is an irreducible polynomial, and its cubic resolvent isIts discriminant of a depressed cubic isFor odd $p$, the $p$-adic valuation of ${\Delta }_p$ is three because $p\nmid 256$; for $p=2$, ${\Delta }_2=2^4\cdot 101$. Thus ${\Delta }_p$ is never a square in $\mathbb{Q}$.

By the rational root theorem, a reducible $g_p$ has an integer root among $\pm 1,\pm p,\pm p^2$. Reduction modulo $p$ shows that the root is divisible by $p$; writing it as $pk$, with $k\in \{\pm 1,\pm p\}$, givesThe only possibilities are $k=-1,p=3$ and $k=1,p=5$, corresponding to roots $-3$ and $5$. For every other prime, $g_p$ is irreducible with nonsquare discriminant, so its Galois group is $S_3$ and the criterion above gives the quartic group $S_4$. For $p=3$ or $5$, the reducible resolvent makes the quartic group a proper subgroup. Therefore

The element $\alpha$ is a root ofBy the rational root theorem, any rational root of this monic polynomial must be one ofDirect substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an irreducible polynomial over $\mathbb{Q}$. It is therefore the minimal polynomial of $\alpha$, and

Irreducibility makes the quartic Galois group a transitive subgroup of $S_4$. Its square discriminant puts it inside $A_4$, leaving $V_4$ or $A_4$. The cubic resolvent isThe rational root theorem shows it is irreducible over $\mathbb{Q}$. Thus the Galois group acts transitively on the three pairings of four roots, excluding $V_4$. Consequently

If a monic polynomial $f$ has roots ${\alpha }_1,\dots ,{\alpha }_n$ in a splitting field, its polynomial discriminant isIt is a symmetric polynomial in the roots, so it belongs to the base field; it is nonzero exactly when $f$ is a separable polynomial.

Putthe Vandermonde determinant. A root permutation $\sigma$ sends $D$ to $\mathrm{sgn}(\sigma )D$. Hence the Galois group of a polynomial is contained in the alternating group $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\mathrm{disc}(f)=D^2$ is a square in $K$. Conversely, if $\mathrm{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the discriminant criterion for an alternating Galois group.

Now letThe rational root theorem shows that none of $\pm 1,\pm 2$ is a root, so this cubic is irreducible over $\mathbb{Q}$. The discriminant of a depressed cubic isThis is not a square in $\mathbb{Q}$, so the Galois group of an irreducible cubic is $S_3$ over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{-19})$, the discriminant is the squareThe cubic remains irreducible: a root in the degree-two extension $K/\mathbb{Q}$ would generate over $\mathbb{Q}$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the tower law for field extensions. Its Galois group over $K$ is consequently $A_3\cong C_3$.