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