Locating zeros of polynomial

Sturm's theorem allows you to do that.

Small example:

$$p(x)=x^2-1/4$$

We compute $$p'(x)=2x$$

and the opposite of the remainder of the division of $p$ by $p'$ which is, up to a positive constant factor, $$1$$

Now we evaluate those three $(p(x),p'(x),1)$ at $x=0$ to get $(-1/4, 0, 1)$ and at $x=1$ to get $(3/4,2,1)$. The first triple has $1$ change of sign, and the second has $0$ changes of sign. Therefore, there are $1-0=1$ zero of $p$ inside $(0,1]$.

If the polynomial $$ (1+x)^np(\frac1{1+x}) $$ has only positive coefficients, then you know that there are no roots in the interval $(0,1)$ by the Descartes rule of signs. If there is one sign change in the coefficient sequence, then there is exactly one root in this interval by the same rule.

Also check the Budan-Fourier theorem, if the number of sign changes in the coefficient sequence of $p(a+x)$ and $p(b+x)$ differs by zero or one, then there are zero or exactly one solutions in $[a,b)$