Discriminant of a trinomial $x^n+ax^m+b$

Question

I am trying to compute the discriminant of the trinomial $x^n+ax^m+b$.

I have tried using resultants but cannot see how to approach it.

Any hints?

Answer 1

$$x^n+ax^m+b=\prod_{k=1}^{n}(x-\zeta_k) \tag{1}$$ $$ n x^{n-1}+am x^{m-1} = \sum_{h=1}^{k}\prod_{\substack{1\leq k\leq n \\ k\neq h}}(x-\zeta_k)\tag{2}$$ and by evaluating $(2)$ at $x=\zeta_i$ we get: $$ n\zeta_i^{n}+am \zeta_i^{m} = \zeta_i\prod_{\substack{1\leq k\leq n \\ k\neq i}}(\zeta_i-\zeta_k)\tag{3} $$ or: $$ (am-an) \zeta_i^{m}-b = \zeta_i\prod_{\substack{1\leq k\leq n \\ k\neq i}}(\zeta_i-\zeta_k)\tag{4} $$ from which: $$ \prod_{i<j}(\zeta_i-\zeta_j)^2 = \pm\frac{1}{b}\prod_{i=1}^{n}\left[(am-an)\zeta_i^m-b\right].\tag{5}$$ If $M$ is the companion matrix of $p(x)=x^n+ax^m+b$, $\{\zeta_i^m\}$ are the eigenvalues of $M^m$, hence the RHS of $(5)$ can be computed from $$ \det\left((am-an)M^m-bI\right) \tag{6}$$