Answer
6 or 4 or 2 or 0 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 $P(x) = x^8 - x^5 + x^4 - x^3 + x^2 - x + 1$
For P(x), there are 6 sign changes. Thus, there are 6 or 4 or 2 or 0 possible positive real roots.
For $P(-x) = (-x)^8 - (-x)^5 + (-x)^4 - (-x)^3 + (-x)^2 - (-x) + 1$
$P(-x) = x^8 + x^5 + x^4 + x^3 + x^2 + x + 1$
There are 0 sign changes. Thus there are 0 real negative real zeros
Thus the answer is 6 or 4 or 2 or 0 real zeros