Today I found a method for finding the roots of a 4th degree polynomial, x4+ax3+bx2+cx+d=0. As with the cubic equation letting x=y-y0 allows us to reduce the number of terms present and get another polynomial of the form y4+By2+Cy+D=0 that can be solved instead. If y is complex then y=p+iq and substitution into the reduced equation results in a 6th degree polynomial for p that is cubic in p2 and its roots can be found using the method for finding the roots of a cubic equation.
Given a set of four roots one can quite easily find the coefficients of the 4th degree polynomial.
The coefficients for the reduced polynomial can then be calculated.
This new set of coefficients allows us to find the coefficients of the cubic equation for p. In this case it has 6 zeros.
With p known we can find the corresponding positive and negative values of q and therefore the y values. Subtracting y0 from each y value gives the corresponding x values for the roots of the original 4th degree polynomial. There is some redundency in the solution but one gets all the original roots even if they are all rational. The reason for this is that q can be imaginary.