How many positive roots can a polynomial have? Descartes gave a simple rule: the number of roots positive is at most the number of sign changes between consecutive coefficients (and, as proved then differs from it by an even number).

See with $x^3 - 6x^2 + 11x - 6$. The signs of the coefficients are $+\ -\ +\ -$: three changes. So there is 3 or 1 positive roots. In fact,

$$x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3),$$

with roots 1, 2 and 3.

For negative roots, change $x$ to $-x$ and count again: $(-x)^3 - 6(-x)^2 + 11(-x) - 6 = -x^3 - 6x^2 - 11x - 6$. The signs are all negative: none exchange, no negative root.

Another example: in $x^3 + x - 2$, the $+\ +\ -$ signals have only one exchange. How the difference has to be Even, there is exactly one positive root, and it is easy to find: $x = 1$.

Test the rule on the three polynomials below. Golden arrows mark each signal change:

Positive roots in gold, negative roots in purple. The last polynomial has 2 exchanges in p(x) and 2 in p(−x).

All topics by René Descartes · ← Square root with ruler and compass · The angles of polyhedra and Euler’s formula →

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