Answer
$6$.
2 or 0 positive real zeros.
2 or 0 negative real zeros.
Work Step by Step
Step 1. Based on the given function, we have degree $n=6$, thus the maximum number of real zeros that the polynomial function may have is $6$.
Step 2. As there are 2 sign changes in $f(x)$, based on the Descartes’ Rule of Signs, there may be 2 or 0 positive real zeros.
Step 3. $f(-x)=2x^6-3x^2+x+1$ and there are 2 sign changes in $f(-x)$, thus there may be 2 or 0 negative real zeros.