This is a question about computing the local residues of a general symmetric bilinear form over $\mathbb{Q}$. I've been using Lam's *Introduction to Quadratic Forms* as a reference, but I'm stuck on the following.

Let $\mathbb{Q}(\alpha)$ be a degree $d$ extension of $\mathbb{Q}$. Given a non-zero element $q=\sum_{i=0}^{d-1}b_i\alpha^i\in\mathbb{Q}(\alpha)$, let $\langle q\rangle\in\mathrm{W}(\mathbb{Q}(\alpha))$ denote the Witt class of the symmetric, non-degenerate bilinear form $\mathbb{Q}(\alpha)\times\mathbb{Q}(\alpha)\to\mathbb{Q}(\alpha)$ given by $(x,y)\mapsto qxy$. Define $\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle\in\mathrm{W}(\mathbb{Q})$ by post-composing $\langle q\rangle$ with the field trace $\mathbb{Q}(\alpha)\to\mathbb{Q}$. For each prime $p$ (including $p=\infty$), we have a local residue map $\partial_p:\mathrm{W}(\mathbb{Q})\to\mathrm{W}(\mathbb{F}_p)$ (or $\mathrm{W}(\mathbb{R})$ for $p=\infty$), and the weak Hasse-Minkowski principle says that the Witt class of $\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ is determined the values $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ for all $p$.

**Question:** Can I compute $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$ in terms of $q$ and the minimal polynomial of $\alpha$ over $\mathbb{Q}$? Or is a general formula too much to ask for?

For example, when $q=1$, this form is a trace form, and I believe a theorem of Taussky-Todd implies that $\partial_\infty\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle 1\rangle\geq 0$ (Corollary I.5.2, *A survey of trace forms of algebraic number fields* by Conner and Perlis).

**Motivation:** I'm writing up an (algebro-geometric) result, and I have a few equations involving terms of the form $\partial_p\operatorname{Tr}_{\mathbb{Q}(\alpha)/\mathbb{Q}}\langle q\rangle$. I can leave them this way, but I'm obviously not a number theorist and would thus like to record these values more explicitly if possible.