Let $M$ be a $4\times 4$ symmetric matrix whose entries $m_{i,j}$ for $i,j =1,\dots,4$ are homogeneous polynomials of degree $2$ in $3$ variables. Assume that $m_{1,1} = 0$.
Does there exist a decomposition of the determinant of $M$ as: $$ \det(M) = A^2 + BC $$ where $A$ is a non zero homogeneous polynomial of degree $4$, and $B,C$ are non constant homogeneous polynomials?
The determinant of $M$ could be written as follows:
$$ (m_{0,1}m_{2,3}+m_{0,2}m_{1,3}+m_{0,3}m_{1,2})^2-2m_{0,1}m_{0,2}m_{2,3}m_{1,3}-4m_{0,1}m_{0,3}m_{1,2}m_{1,3}-4m_{0,2}m_{0,3}m_{1,2}m_{1,3}+2m_{0,1}m_{0,2}m_{1,2}m_{3,3}+2m_{0,2}m_{0,3}m_{1,1}m_{2,3}-m_{0,1}^2m_{2,2}m_{3,3}-m_{0,2}^2m_{0,3}m_{1,1}-m_{0,3}^2m_{1,1}m_{2,2}. $$
So, if for instance $m_{0,1}=0$, the answer is positive since we have $$ \det(M) = (m_{0,2}m_{1,3}+m_{0,3}m_{1,2})^2 +m_{0,3}(-4m_{0,2}m_{1,2}m_{1,3}+2m_{0,2}m_{1,1}m_{2,3}-m_{0,2}^2m_{1,1}-m_{0,3}m_{1,1}m_{2,2}). $$
