Let's say you have the polynomial with fractional power: $$P(z)=z^{9/2}-\frac{3}{2} z^4 +\frac{1}{2} z^3=0, \quad z \in \mathbb{C} \tag{1}$$ I have found two real roots of Eq. (1): $$z_1=0,$$ and $$z_2=1.$$
I'm looking for a theorem that define the number of roots and their multiplicity in the interval $[-1, 1]$.
I have found the Sturm's theorem that expresses the number of distinct real roots of $P(x)$ located in an interval. Also I have read the Budan's theorem, is the theorem for computing an upper bound on the number of real roots a polynomial has inside an open interval.
The Sturm's theorem and the Budan's theorem are both for an integer power of real polynomial equation $P(x)$, $x \in \mathbb{R}$. Later I found bounds on (complex) polynomial roots based on the Rouche's theorem that is usually used to simplify the problem of locating zeros.
In my case the highest power is fractional, $\frac{9}{2} \in \mathbb{R}$.
Question.
Does the rule exist to find or estimate the
lower/upper bound on the number of roots (real or comlex) a polynomial in an interval when polynomial's power is not integer?
Edit. If I wrote: solve x^(9/2)-3/2*x^4 +x^3/2 then Wolframalpha gives two roots only: 0 and 1.
In this specific way, we can make a substitution $u = z^{1/2}$ to end up with $$ \begin{split} 0 &= u^9-\frac32u^8+\frac12u^6\\ &= \frac{u^6}{2}\left(2u^3-3u^2+1\right)\\ &= \frac{u^6}{2} (u-1) \left(2u^2-u-1\right)\\ &= \frac{u^6}{2} (u-1)^2 \left(2u+1\right).\\ \end{split} $$ So there are $9$ roots: $6$ multiplicities of $u_1=0$, $2$ multiplicities of $u_2=1$, and $u_3=-1/2$. So there must be $18$ roots of the original equation.
This suggests the approach for the rational powers -- substitute $u=z^{1/n}$ where $n$ is what makes all remaining powers integers in the new equation in $u$, and apply the Fundamental Theorem of Algebra to get your upper bound.