## 1. Graphical solution of the cubic equation

Graphical solution of the cubic equation   a3x3+a2x2+a1x+a0 = 0.
The procedure of definition of the polygon p0 = ABCDE with four sides, each  pi/2-rotated with respect to the previous one, is described in GraphicalSolution.html . Here the process is repeated for the case of the cubic equation starting with a segment AB of oriented length a3, proceeding to one orthogonal to it BC of length a2 and so on. The graphical solution is achieved by finding another polygon p1 constructed analogously and satisfying the two conditions.
1) p1 starts at A and ends at E.
2) p1 is inscribed in p0.
The solution corresponding to p1 is given by.
xi = -tan(fi).
The proof is again assured by verifying the Horner schema.
(   (a3(-tan(fi))+a2) * (-tan(fi))+a1)   )*(-tan(fi)) + a0  = 0.

The figure displays the auxiliary polygon p2=AB'C'FE which reduces to p1 when E and F coincide.
p2 is constructed as follows.
1) B' is arbitrary on BC (2nd side of p0).
2) C' is the intersection of CD (3rd side of p0) with the orthogonal B'C' to AB'.
3) C'F is orthogonal to B'C' at C' and F is the projection of E on C'F.
Curve c is the geometric locus of F as B' varies on BC. It passes so many times through E as is the number of real solutions of the equation.
a3x3+a2x2+a1x+a0=0.
Depending on the coefficients {ai} there are one two or three places for B' on BC for which E and F coincide. For each such place we obtain a corresponding solution of the cubic through x = -tan(fi).

## 2. Related questions

Some questions connected with this subject.
1) Given the polygon p0 = ABCDE determine the polygon p1. This amounts in this case to      finding a direction e, so that projecting points {A, E} parallel to e onto B' on BC and      onto C' on DC line B'C' is orthogonal to e.
2) Polygon p1 can be used in the same way to find further real roots of the original polynom.
In fact it is easily proved that p1 represents the polynomial resulting by dividing the original     a3x3+a2x2+a1x+a0   by  the factor  x-x0, where x0 is the root (x0 = -tan(fi))     corresponding to p1. This is valid vor any degree polynomial.
3) Find the equation of the locus described by point F. Do the same for the more general polygon p2     resulting from polygons of any degree n.
4) A locus problem related to this discussion is studied in MaclaurinDual.html .
5) The corresponding construction for the quartic is discussed in GraphicalSolutionQuartic.html .