I recently started with the polynomial $$P(z) = z^4 + bz^3 + cz^2 + dz + e$$ and calculated an expression closely related to the discriminant of the polar derivative of this polynomial. This expression was a sum of a number of nasty degree $6$ polynomials which, after a few of these polynomials were omitted in a natural way, resulted in the rest of them having their higher power terms cancel out. This yielded the following two (still somewhat daunting) polynomials of degrees $3$ and $2$, respectively:
\begin{align} Q(z) & = -4\left(b^3-4bc+8d\right)\left(27b^2e-9bcd+2c^3-72ce+27d^2\right)z^3 \\ & -4\left(27b^4ce+99b^4d^2-75b^3c^2d-144b^3de+13b^2c^4-114b^2c^2e-453b^2cd^2-360 b^2e^2+326bc^3d\right. \\ & \left.+1032bcde+648bd^3-52c^5-32c^3e-306c^2d^2+960ce^2-1944d^2e\right)z^2 \\ & + 4\left(-324b^4de+99b^3c^2e+42b^3cd^2+972b^3e^2-13b^2c^3d+1176b^2cde+72b^2 d^3-368bc^3e-201bc^2d^2\right. \\ &\left.-3648bce^2-2196bd^2e+52c^4d+768c^2de+108cd^3+6336de^2\right)z \\ &+2835b^4e^2-2322b^3cde+568b^3d^3+560b^2c^3e-153b^2c^2d^2-13824b^2ce^2+654b^2d^2 e+9600bc^2de\\ &-2394bcd^3+14592bde^2-2096c^4e+576c^3d^2+11136c^2e^2-14400cd^2e+3051d^4-11008e^3 \end{align}
\begin{align} R(z) & = 4\left(3 b d-c^2-12 e\right) \left(3 b^3 d-b^2 c^2+6 b^2 e-14 b c d+4 c^3-16 c e+18 d^2\right)z^2 \\ & + 4 \left(3 b d-c^2-12 e\right) \left(9 b^3 e-b^2 c d-32 b c e-3 b d^2+4 c^2 d+48 d e\right)z \\ & -6453 b^4 e^2+4338 b^3 c d e-968 b^3 d^3-968 b^2 c^3 e+243 b^2 c^2 d^2+34272 b^2 c e^2-1434 b^2 d^2 e-19200 b c^2 d e \\ & +4338 b c d^3-45312bde^2+3856 c^4 e-968 c^3 d^2-30336 c^2 e^2+34272 c d^2 e-6453d^4+59648e^3 \end{align}
At first it looked hopeless to figure out where these expressions in the coefficients of $P$ came from. Eventually, however, I found that the two factors $\left(b^3-4bc+8d\right)$ and $\left(27b^2e-9bcd+2c^3-72ce+27d^2\right)$ in the coefficient of $z^3$ in $Q(z)$, and the factor $(3bd-c^2-12e)$ in the coefficients of $z$ and $z^2$ in $R(z)$ apparently show up in Ferrari's/Cardano's general solution of the quartic equation; see here (and click on "Exact forms").
I have a number of reasons to believe that these polynomials $Q(z)$ and $R(z)$ can be simplified in very elegant ways, but I've been stuck on this problem for several days after my above discovery. Any assistance, or ideas about more general contexts in which these terms in Ferrari's formula may show up, would be greatly appreciated.