Answer
1 positive real zeros.
2 or 0 negative real zeros.
Work Step by Step
Step 1. There is 1sign change in $f(x)$, based on the Descartes’ Rule of Signs, there will be 1 positive real zeros.
Step 2. $f(-x)=6x^5+x^4-5x^3-x+1$, there are 2 sign changes in $f(-x)$, based on the Descartes’ Rule of Signs, there may be 2 or 0 negative real zeros.
