I would like to factor the quartic into two quadratic polynomials $F(g),G(h)$: \begin{align*} x^{4}+ax^{3}+bx^{2}+cx+d & =(x^{2}+g_{1}x+h_{1})(x^{2}+g_{2}x+h_{2}),\\ & =x^{4}+(g_{1}+g_{2})x^{3}+(g_{1}g_{2}+h_{1}+h_{2})x^{2}+(g_{1}h_{2}+g_{2}h_{1})x+h_{1}h_{2}, \end{align*} which leads to the following conditions (C1,C2,C3,C4) between the roots $g_{1},g_{2}$ and $h_{1},h_{2}$ \begin{align*} g_{1}+g_{2} & =a,\\ g_{1}g_{2}+h_{1}+h_{2} & =b,\\ g_{1}h_{2}+g_{2}h_{1} & =c,\\ h_{1}h_{2} & =d. \end{align*} of the quadratic polynomials $F(g)$ and $F(h)$ \begin{align*} F(g) & =g^{2}+Ag+B,\\ G(h) & =h^{2}+Ch+D \end{align*} Using the Vieta's theorem, it obvious that 2 coefficients may be determined directly $A=-a$, and $D=d$. However, the remaining coefficients $B,C$ look more difficult. In the paper
https://ijpam.eu/contents/2011-71-2/7/7.pdf
I found the following solution. These quadratic polynomials \begin{align*} F(g,y) & =g^{2}-ag+b-y=0,\\ G(h,y) & =h^{2}-yh+d=0, \end{align*} hold C1,C2,C4, their roots are \begin{align*} g_{1,2} & =\frac{a}{2}\pm\frac{\sqrt{a^{2}-4b+4y}}{2},\\ h_{12} & =\frac{y}{2}\pm\frac{\sqrt{y^{2}-4d}}{2}. \end{align*} Using the C3 condition, $y$ is determined from \begin{align*} g_{1}h_{2}+g_{2}h_{1} & =\left(\frac{a}{2}+\frac{\sqrt{a^{2}-4b+4y}}{2}\right)\left(\frac{y}{2}-\frac{\sqrt{y^{2}-4d}}{2}\right)+\left(\frac{a}{2}-\frac{\sqrt{a^{2}-4b+4y}}{2}\right)\left(\frac{y}{2}+\frac{\sqrt{y^{2}-4d}}{2}\right)=c\\ & =\frac{ay-\sqrt{y^{2}-4d}\sqrt{a^{2}-4*b+4*y}}{2}=c, \end{align*} which leads to the cubic equation for $y$ $$ y^{3}-by^{2}+(ac-4d)y+4bd-da^{2}-c^{2}=0. $$ Let us do the following example: $a=-0.25,$ $b=-0.85$, $c=1.45$, $d=-4.35$. There are two complex and 1 real roots of the cubic
y1=-0.043202328926409 + 4.119424154478284i
y2=-0.043202328926409 - 4.119424154478284i
y3=-0.763595342147182
The roots of $F(g,y)$ are real
g1=-0.444420816248437, g2= 0.194420816248437
as well as for the $G(h,y)$
h1=-2.502120632708732, h2 = 1.738525290561551
The back substitution leads to \begin{align*} g_{1}+g_{2} & =-0.25,\\ g_{1}g_{2}+h_{1}+h_{2} & =-0.85,\\ g_{1}h_{2}+g_{2}h_{1} & =-1.259101164463204,\\ h_{1}h_{2} & =-4.35. \end{align*} Unfortunately, the relationship C3 is not held even for the complex roots. This issue affects the quartic roots, which are incorrect:
\begin{align*} x_{12} & =\frac{1}{2}(-g_{1}\pm\sqrt{g_{1}^{2}-4h_{1}}),\\ x_{34} & =\frac{1}{2}(-g_{2}\pm\sqrt{g_{2}^{2}-4h_{2}}). \end{align*}
Where is the problem? An incorrect formulas for $F(g,y)$ or $G(h,y)$?
Thanks for your help.
The Matlab code:
clc
clear
syms a b c d y g h
format long
%Roots of quadratics
g = solve(g^2 - a*g + b - y, g)
h = solve(h^2 - y*h + d, h)
g1 = g(1); g2 = g(2);
h1 = h(1); h2 = h(2);
%Verifying the conditions
C1 = g1 + g2; %OK
C2 = simplify(g1 * g2 + h1 + h2); %OK
C4 = simplify(h1 * h2); %OK
%Cubic for C3
C3 = simplify(g1 * h2 + g2 *h1);
Q = expand(((y^2 - 4*d)*(a^2 - 4*b + 4*y))/4-((a*y)/2 -c)^2);
%Numerical verification
a = -0.25; b=-0.85; c = 1.45; d =-4.35;
%Roots of cubic and booth quadratics
y = roots([1 -b (a*c - 4*d) (-a^2*d - c^2 + 4*b*d)]);
y = y(3); %Use the real root
g = roots([1 -a (b - y)]);
h = roots([1 -y d]);
%Check the conditions
C11 = g(1) + g(2); %OK
C21 = g(1) * g(2) + h(1) + h(2); %OK
C31 = g(1) * h(2) + g(2) *h(1); % WRONG !
C41 = h(1) * h(2); %OK
%Roots of the quartic
x1 = 0.5*(-g(1) + sqrt(g(1)^2-4*h(1)))
x2 = 0.5*(-g(1) - sqrt(g(1)^2-4*h(1)))
x3 = 0.5*(-g(2) + sqrt(g(2)^2-4*h(2)))
x4 = 0.5*(-g(2) - sqrt(g(2)^2-4*h(2)))
