Answer
$f(x)$ has $1$ positive zeros. and $f(x)$ has $2$ or $0$ negative zeros
Work Step by Step
Descartes’ Rule of Signs
Let $f$ denote a polynomial function written in standard form.
The number of positive real zeros of $f$ either equals the number of variations in the sign of the nonzero coefficients of $f(x)$ or else equals that number less an even integer.
The number of negative real zeros of $f$ either equals the number of variations in the sign of the nonzero coefficients of $f(-x)$ or else equals that number less an even integer
Therefore, $f(x)=-6x^5+x^4+5x^3+x+1,$ has $1$ positive zeros. and $f(-x)=6x^5+x^4-5x^3-x+1$ has $2$ or $0$ negative zeros
