I want to show it is possible to choose $m$ such that $x^p + mp^2 \left(x^{p-2} + \sum_1^{p-2} (-1)^i e_i x^{p-2-i} \right) -p$ has exactly $p-2$ real roots, where $e_i$ are positive constants. (So we really just care that all $e_i > 0$) It should be noted that $p$ is prime.
My approach has been via Sturm's Theorem, but this is extremely painful as it involves calculating many derivatives and carrying out lots of polynomial division.
My question is whether there exists a more elegant approach to this calculation or an observation that I have missed.
When we expand the polynomial we can see the coefficients on $x^k$ are positive for $k$ odd and negative for $k$ even.
Using Descarte's rule of signs we see the sequence: $$+,\underbrace{+,-,+,-,\dots}_{p-2 \text{ times}}$$ So we know the number of positive roots are at most $p-2$. Using the transform $x \to -x$ yields that there are no negative roots. Thus by Fundamental Theorem of Algebra we must have at least two complex roots. We can conclude with the method demonstrated in this answer: Existence of irreducible polynomial over $\mathbb Q$, which has exactly two non-real roots (Using Rouche's Theorem)