Answer
$3$.
1 positive real zero.
2 or 0 negative real zeros.
Work Step by Step
Step 1. Based on the given function, we have degree $n=3$, thus the maximum number of real zeros that the polynomial function may have is $3$.
Step 2. As there is 1 sign change in $f(x)$, based on the Descartes’ Rule of Signs, there will be 1 positive real zero.
Step 3. $f(-x)=x^3-x^2-x+1$ and there are 2 sign changes in $f(-x)$, thus there may be 2 or 0 negative real zeros.