Roots of a Degree 7 Polynomial

The quantity $Q(a)$ is known as the resultant $R(f_a,g).$ The resultant of two polynomials is the product of the evaluations of the second polynomial on the roots of the first polynomial (counted with multiplicities). It has several nice properties, two of which are:

(1) $R(F,G) = \pm R(G,F)$ (in fact the sign is negative iff the degrees of $a,b$ are odd)

(2) If $F=GQ+S$ and $G$ is monic, $R(F,G) = R(S,G).$

The proofs are left as exercises; both are quite straightforward.

This gives an algorithm for computing resultants in general, and for this particular case, we only need property (2): $R(f_a,g) = R(X+a-3,g) = g(3-a) = -a^3+9a^2-30a+33.$ So the answer is $\fbox{73}.$

Note first that $f_a(X) \; = \; (X^4 + 2X^2 - X - 1)g(X) + X + a-3$ If the zeroes of $g(X)$ are $u_1,u_2,u_3$ , then we obtain $\begin{aligned} Q(a) & = \; \prod_{j=1}^7 g(\alpha_{a,j}) \; = \; \prod_{j=1}^7 \prod_{k=1}^3 (\alpha_{a,j} - u_k) \; = \; \prod_{k=1}^3 \prod_{j=1}^7 (\alpha_{a,j} - u_k) \\ & = \; -\prod_{k=1}^3 f_a(u_k) \; = \; -\prod_{k=1}^3 \big(u_k + a - 3\big) \; = \; \prod_{k=1}^3 \big(3-a - u_k\big) \\ & = \; g(3-a) \; = \; 33 - 30a + 9a^2 - a^3 \end{aligned}$ so that $|A| + |B| + |C| + |D| = 1+9+30+33 = \boxed{73}$ .