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Consider the polynomial $p(x)= x^3-x-1$.

Descartes' rule of sign:
Looking at the signs of the coefficients of $p(x): + - -.$ Therefore the polynomial must have exactly 1 positive real root.

Looking at the signs of the coefficients of $p(-x): - + -.$ Therefore the polynomial must have $2$ or $0$ negative real roots.

So it follows the polynomial $p(x)$ has $2$ or $0$ complex roots. Is there a way (without explicitly finding the roots) to give the amount of complex roots?

(the reason for this question: To calculate the Minkowski bound of $\mathbb{Q}(\alpha)$ with $\alpha$ a root of $p(x)$ it is necessary to know the amount of complex roots.)

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Yes: compute its discriminant and check its sign.

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Yes. Calculate the derivative of $p$, find the roots of the derivative, and if those are real, check the signs of $p$ at these local extrema. If the minimum is negative and the maximum is positive, then you have three real roots, and if they have the same sign (or there are no local extrema) you have a single real root. If any of them are $0$, you have a multiple real root.

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If $x < -1$ then $x^3 - x < 0$ and $x^3 - x - 1 < 0$

And if $-1 \le x \le 0$ then $-x -1 \le 0$ and $x^3 - x - 1 < 0$

There are no roots less than $0.$

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