# Solving the general fourth degree equation

Let $f=X^{4}+aX^{3}+bX^{2}+cX+d \in K[X]$ where $K[X]$ is a field with $char(K) \neq 2,3$ and let $\alpha_{1}, \dots , \alpha_{4}$ be its roots( in an extension of $K$ ). Define $$C_{1}=(\alpha_{1}+\alpha_{2}-\alpha_{3}-\alpha_{3})^{2}$$ $$C_{2}=(\alpha_{1}-\alpha_{2}+\alpha_{3}-\alpha_{4})^{2}$$ $$C_{3}=(\alpha_{1}-\alpha_{2}-\alpha_{3}+\alpha_{4})^{2}$$ The exercise at hand first asks to verify that the $S_{4}$ action permutes the $C_{i}$ and that the subgroup $H=\{ (1),(12)(34),(13)(24),(14)(23) \}$ fixes the $C_{i}$ and then to show that the following relations hold: $$C_{1}+C_{2}+C_{3}=3a^{2}-8b$$ $$C_{1}C_{2}+C_{2}C_{3}+C_{1}C_{3}=3a^{4}-16a^{2}b+16b^{2}+16ac-64bd$$ $$C_{1}C_{2}C_{3}=(a^{3}-4ab+8c)^{2}$$ This were clearly just computational and I solved them quickly, having used Newton's identities for instance for the second task. I couldn't provide an answer though for the last question regarding this exercise which is to verify that, given the above information, one can solve in general the fourth degree equation.

What I did was to express $\alpha_{1}$ in terms of the $\sqrt{C_{i}}'$s and the coefficient $a$ of $f$. Then I think one could get the other roots by solving the system of equations in the square roots of the $C_{i}'$s which if I am not wrong means solving a third degree equation. I am quite sure this is not the way to do it though.

In the literature, I have read about solving fourth degree equations in general, including the wikipedia article regarding this and though they seem to have a similar approach to the one this exercise provides, I wasn't able to deduce how to solve the general fourth degree equation using only the information provided by this exercise.

I would appreciate any help concerning this. Thank you!

• Expand $(x-C_1)(x-C_2)(x-C_3)$. What are the coefficients? – tehjh Mar 13 '17 at 15:31
• Well, of course the coefficients will be the elementary symmetric polynomials in the $C_{i}$'s which we now how to express them in terms of the coefficients of f, but how does one actually get the roots? – Raizen Mar 14 '17 at 12:49
• Ah yes sorry I did not see that you had already arrived at a third degree equation, which is in fact right, because the solving of a cubic is well known: first depress it and then the most elegant way would be through factorizing $x^3+p^3+q^3-3xpq$ and looking at it as a polynomial in $x$ – tehjh Mar 14 '17 at 12:53