# Number of roots in interval by Fourier

To bound from above the numbers of polynomial roots in the open interval, one can employ Fourier's theorem: evaluate each of the interval's endpoints over the given polynomial and all of its derivatives. Count number of times the signs in the resulting sequences changed, that's the number of roots (0 or 1) or the the upper bound otherwise.

For example:

$$x^2 - 3x + 2 = (x-1)(x-2)$$

Derivatives are $2x - 3$ and $2$.

Evaluated over open inteval $(1, 2)$ results in:

$$0, -1, 2$$

$$0, 1, 2$$

So the first sequence has one sign variation, the other has zero. That means there should be exactly one root.

Please, I have two questions: as the roots are $1$ and $2$, why does it says that in the open interval $1, 2$ there is more than zero roots? And why then does it say there is one root and not two, and which of the two roots does it report?

I think the problem is, that the right end of your interval is a root of the polynomial. Wikipedia says "... provided $p(r)\neq 0"$, the following applies: ..."
So if you take an interval $(1,r)$ with an $r>2$, you get 1 for the number of roots. And this is the correct number because $2\in (1,r)$.