Quartic formula

Remarks

Since $q \ne 0$, the constant term $-q^2$ of the resolvent cubic is nonzero, so $m \ne 0$. The resolvent cubic says exactly that $(y^2 + \frac{p}{2} + m)^2 - (y^4 + py^2 + qy + r) = 2my^2 - qy + (m^2 + pm + \frac{p^2}{4} - r)$ is the square $(sy - \frac{q}{2s})^2$. So the quartic factors as $(y^2 + sy + \frac{p}{2} + m - \frac{q}{2s})(y^2 - sy + \frac{p}{2} + m + \frac{q}{2s})$, and the quadratic formula gives the four roots. The resolvent cubic itself is solved by the cubic formula, so all roots are radical expressions in the coefficients. If $q = 0$, the equation $y^4 + py^2 + r = 0$ is a quadratic in $y^2$, and $y^2 = \frac{-p \pm \sqrt{p^2 - 4r}}{2}$. Example: for $y^4 - 2y^2 + 8y - 3$, the value $m = 2$ is a root of $8m^3 - 16m^2 + 32m - 64$, so $s = 2$ and the factors are $y^2 + 2y - 1$ and $y^2 - 2y + 3$.

Going deeper

For a student

This method is due to Ferrari, and it shows that every quartic is solvable by radicals. In Galois terms, $S_4$ is solvable, with the normal subgroup $V_4$ of order $4$ and quotient $S_3$. If $y_1, y_2, y_3, y_4$ are the roots, each root $m$ of the resolvent cubic has the form $\frac{1}{2}(y_i + y_j)^2$, and $(y_1 + y_2)^2$ is fixed by $V_4$. This is why a cubic appears. For degree $5$ no such reduction exists, by the Abel–Ruffini theorem.