Skip to main content
corrected formula
Source Link

If I did not mess up the adaptation to your special case, Theorem 3.31 for $n=2$ in "A computational approach to the discriminant of homogeneous polynomials" by Busé and Jouanolou says that $$ {\rm Disc}(R)=({\rm Disc}(F))^r\times {\rm Res}(A,B)^{d(d-1)}\times K(F,A,B) $$ where $K$ is homogeneous of degree $2dr(r-1)$$2d(r-1)$ in the coefficients of $A$ or $B$ and of degree $2(r-1)$ in the coefficients of $F$.

If I did not mess up the adaptation to your special case, Theorem 3.31 for $n=2$ in "A computational approach to the discriminant of homogeneous polynomials" by Busé and Jouanolou says that $$ {\rm Disc}(R)=({\rm Disc}(F))^r\times {\rm Res}(A,B)^{d(d-1)}\times K(F,A,B) $$ where $K$ is homogeneous of degree $2dr(r-1)$ in the coefficients of $A$ or $B$ and of degree $2(r-1)$ in the coefficients of $F$.

If I did not mess up the adaptation to your special case, Theorem 3.31 for $n=2$ in "A computational approach to the discriminant of homogeneous polynomials" by Busé and Jouanolou says that $$ {\rm Disc}(R)=({\rm Disc}(F))^r\times {\rm Res}(A,B)^{d(d-1)}\times K(F,A,B) $$ where $K$ is homogeneous of degree $2d(r-1)$ in the coefficients of $A$ or $B$ and of degree $2(r-1)$ in the coefficients of $F$.

Source Link

If I did not mess up the adaptation to your special case, Theorem 3.31 for $n=2$ in "A computational approach to the discriminant of homogeneous polynomials" by Busé and Jouanolou says that $$ {\rm Disc}(R)=({\rm Disc}(F))^r\times {\rm Res}(A,B)^{d(d-1)}\times K(F,A,B) $$ where $K$ is homogeneous of degree $2dr(r-1)$ in the coefficients of $A$ or $B$ and of degree $2(r-1)$ in the coefficients of $F$.