Answer
3 or 1 real zeros
Work Step by Step
For Descartes' Rule of Signs, the number of sign changes (+ then - or - then +) for P(x) will be the number of possible positive real roots. The number of sign changes for P(-x) will be the number of possible negative real roots.
Given $P(x) = x^3 - x^2 - x - 3$
For P(x), there is one sign change. Thus, there is 1 possible positive real root.
For $P(-x) = (-x)^3 - (-x)^2 - (-x) - 3$
$P(-x) = -x^3 - x^2 + x - 3$
There are two sign changes. Thus there is 2 or 0 negative real roots.
Thus the answer is either 3 or 1 real zeros
