Vtyjkyuk

I would like to factor the quartic into two quadratic polynomials
$F(g),G(h)$:
beginalign*
x^4+ax^3+bx^2+cx+d & =(x^2+g_1x+h_1)(x^2+g_2x+h_2),\
& =x^4+(g_1+g_2)x^3+(g_1g_2+h_1+h_2)x^2+(g_1h_2+g_2h_1)x+h_1h_2,
endalign*
which leads to the following conditions (C1,C2,C3,C4) between the
roots $g_1,g_2$ and $h_1,h_2$
beginalign*
g_1+g_2 & =a,\
g_1g_2+h_1+h_2 & =b,\
g_1h_2+g_2h_1 & =c,\
h_1h_2 & =d.
endalign*
of the quadratic polynomials $F(g)$ and $F(h)$
beginalign*
F(g) & =g^2+Ag+B,\
G(h) & =h^2+Ch+D
endalign*
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
beginalign*
F(g,y) & =g^2-ag+b-y=0,\
G(h,y) & =h^2-yh+d=0,
endalign*
hold C1,C2,C4, their roots are
beginalign*
g_1,2 & =fraca2pmfracsqrta^2-4b+4y2,\
h_12 & =fracy2pmfracsqrty^2-4d2.
endalign*
Using the C3 condition, $y$ is determined from
beginalign*
g_1h_2+g_2h_1 & =left(fraca2+fracsqrta^2-4b+4y2right)left(fracy2-fracsqrty^2-4d2right)+left(fraca2-fracsqrta^2-4b+4y2right)left(fracy2+fracsqrty^2-4d2right)=c\
& =fracay-sqrty^2-4dsqrta^2-4*b+4*y2=c,
endalign*
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
beginalign*
g_1+g_2 & =-0.25,\
g_1g_2+h_1+h_2 & =-0.85,\
g_1h_2+g_2h_1 & =-1.259101164463204,\
h_1h_2 & =-4.35.
endalign*
Unfortunately, the relationship C3 is not held even for the complex
roots. This issue affects the quartic roots, which are incorrect:

beginalign*
x_12 & =frac12(-g_1pmsqrtg_1^2-4h_1),\
x_34 & =frac12(-g_2pmsqrtg_2^2-4h_2).
endalign*

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)))