Sum of negative roots of a $5^{th}$ degree monic polynomial Let $f(x)$ be a $4^{th}$ degree monic polynomial say $f(x) = x^4 + a_1x^3+a_2x^2+a_3x+a_4$ with the property that $a_1<0, a_4>0$ and $a_2<a_3$. They by Descartes' rule of signs we can conclude that there are $2$ positive roots and $2$ negative roots of $f(x) = 0$.
$\textbf{Question 1:}$ Is there any similar result for $5^{th}$  degree monic polynomial, i.e. with the given coefficients can we have some relationships that will tell us there are exactly $2$ positive roots and $3$ negative roots?
My intuition: Let $g(x) = x^5+ b_1x^4 + b_2x^3+b_3x^2+b_4x+b_5$ with $b_1<0, b_5>0$ and $b_2<b_3<b_4$ we can say something silimar (I am not sure about this).
$\textbf{Question 2:}$ Suppose $g(x)$ be a $5^{th}$  degree monic polynomial. Can we determine the sum of negative roots of $g(x) = 0$ given the coefficients? Is there any existing literature that may help me in this regard?
Thank you very much!!
 A: You can answer these questions using quantifier elimination. Here's how to do it for cubics in Mathematica:
To find out whether a monic cubic has two negative roots, Mathematica takes the input
f[u_] := u^3 + a u^2 + b u + c
Reduce[Exists[{v, w, x}, v < w < 0 < x &&
   f[v] == f[w] == f[x] == 0 ], {a, b, c}, Reals]

The output is that this occurs when
\begin{align}
\big(a\leq 0\land b<0\phantom{^2} \land\ & 9 a b-2 a^3 -2 \sqrt{D}<27c<0\big)\\
\lor\ \big(a>0 \land b<\frac{a^2}{4}\land\ &
   9 a b-2 a^3 -2 \sqrt{D}<27c<0\big)
\end{align}
where $D=a^6-9a^4 b+27 a^2 b^2-27 b^3$.
To find a formula for the sum of the two negative roots, Mathematica takes the input
f[u_] := u^3 + a u^2 + b u + c
Reduce[Exists[{v, w, x}, v < w < 0 < x && v + w == y &&
  f[v] == f[w] == f[x] == 0], {a, b, c, y}, Reals]

The output is that this occurs when the above inequalities hold and $y$ is the first root of
$$y^3 + 2a y^2 + (a^2+b)y + ab -c =0$$
One could also express this $y$ in terms of square roots and cube roots.
These are only slighty simplified from the raw Mathematica output. You could get similar results for quintics too, with a longer process and longer expressions.
