Consider the polynomials $$p_1=3(56-56A+A^2)-112(-6+A)S-14(-36+A)S^2+112S^3+7S^4,$$ $$p_2=-112(-6+A)-28(-36+A)S+336S^2+28S^3,$$ $$p_3=-28(-36+A)+672S+84S^2,$$ $$p_4=672+168S,$$ $$p_5=168,$$ $$q=-168S-252S^2-84S^3-7S^4+84A+84AS-3A^2+14AS^2.$$ I want to prove that if $A>0$, $S<0$ and $p_i>0$ then $q>0$.
I have observed that $\frac{\partial p_i}{\partial S}(A,S)=p_{i+1}(A,S)$ and $q=-p_1+p_2-p_3+p_4-p_5$, so we can express $q$ as $$q=\sum_{k=0}^4(-1)^{k+1}\frac{\partial^k p_1}{\partial S^k},$$ but I don't know if it can be used to prove the above statement.