Answer
$3$.
2 or 0 positive real zeros.
1 negative real zero.
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 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. $f(-x)=-3x^3-2x^2-x+2$ and there is 1 sign change in $f(-x)$, thus there will be 1 negative real zero.
