so I can assume the B1 and B2 are diagonal and then algebraically there's no difference if I assume the whole B1 and B2 block is just 1x1, so M becomes x2^2 + (2x1+Ax2+Bx3)x3. The extra term (2x1+Ax2+Bx3)x3 always contributes 1-1=0 to the signature, using the trick y1y2 = (1/4)[(y1+y2)^2-(y1-y2)^2].
e.g. x2^2 + 2x1x3 + 18x2x3 + x3^2 = x2^2 + (x3+9x2+x1)^2 - (9x2+x1)^2