Representing polynomials as quadratic forms I have the following cubic equation
$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3)=0$$
whose solutions are $x=1,2,3$.
What if the above equation were represented by quadratic form? Let $\mathbf x = \begin{bmatrix} x^2 & x & 1\end{bmatrix}^T$. Can the cubic equation above be represented as follows?
$$x^3-6x^2+11x-6=\begin{bmatrix}x^2&x&1\end{bmatrix}\begin{bmatrix}0&b&c\\d&e&f\\g&h&-6\end{bmatrix}\begin{bmatrix}x^2\\x\\1\end{bmatrix} = \textbf{x}^T \textbf{A} \textbf{x}$$
There are many possibilities to form matrix $A$. For instance the two matrices below
$$\textbf{A}_1 = \begin{bmatrix}0&0.5&0\\0.5&-6&5.5\\0&5.5&-6\end{bmatrix}\quad\quad\quad\textbf{A}_2 = \begin{bmatrix}0&0&0\\1&-3&6\\-3&5&-6\end{bmatrix}$$
Is there any study about finding the homogeneous solution, $\textbf{x}^T \textbf{A} \textbf{x}=0$? Thank you in advance.
 A: The simplest solution is to use the 3 homogeneous coordinates $\pmatrix{ x^2 & x & 1}$ and a 3×3 symmetric matrix
$$A x^3 + B x^2 + C x + D = \pmatrix{x^2 \\x \\1}^\top 
\begin{bmatrix} 
  0 & \frac{A}{2} & 0 \\ 
  \frac{A}{2} & B & \frac{C}{2} \\ 
  0 & \frac{C}{2} & D 
\end{bmatrix} \pmatrix{x^2 \\x \\1} $$
But the formal solution is actually to use 4 homogeneous coordinates $\pmatrix{ x^3 & x^2 & x & 1}$ and a 4×4 symmetrix matrix
$$A x^3 + B x^2 + C x + D = \pmatrix{x^3 \\ x^2 \\x \\1}^\top \begin{bmatrix} 
  0 & 0 & 0 & 0\\
  0& 0 & \frac{A}{2} & 0 \\ 
  0 & \frac{A}{2} & B & \frac{C}{2} \\ 
  0 & 0 & \frac{C}{2} & D 
\end{bmatrix} \pmatrix{x^3 \\ x^2 \\x \\1} $$
The reason is that you have to treat $x^3$ as a separate variable from $x^2$, $x$ and $1$. The coefficients are found by matching and taking successive derivatives of both sides of the expression. 
Additional constraints I have imposed (besides the matrix being symmetric) is to make it as diagonal as possible, by making the most off-diagonal terms zero.
A: With $\mathbb x=[x,y,1]^T$, the equation $\mathbb x^TA\mathbb x=0$ defines a conic, which corresponds to a cone in homogeneous coordinates $\mathbb x=[x,y,z]^T$.
Now when you consider the curve $\mathbb x=[x^2,x,1]^T$, you have the parabola $x=y^2$ in the plane of the conic.
The intersection of the parabola and the conic is made of up to four points and this is no surprise as $[x^2,x,1]A[x^2,x,1]^T=0$ defines a quartic equation.
Presumably, in the case the quartic degenerates to a cubic, the parabola tangents the conic.
