Answer
$f(x)$ has $3$ or $1$ positive roots.
$f(x)$ has no negative roots
Work Step by Step
Positive roots
The rule states that if the nonzero terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign changes between consecutive (nonzero) coefficients, or is less than it by an even number. A root of multiplicity k is counted as k roots.
In particular, if the number of sign changes is zero or one, the number of positive roots equals the number of sign changes.
Negative roots
As a corollary of the rule, the number of negative roots is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than it by an even number. This procedure is equivalent to substituting the negation of the variable for the variable itself. For example, the negative roots of $ax^3+bx^2+cx+d$ are the positive roots of
${\displaystyle a(-x)^{3}+b(-x)^{2}+c(-x)+d=-ax^{3}+bx^{2}-cx+d.}$
Thus, applying Descartes' rule of signs to this polynomial gives the maximum number of negative roots of the original polynomial.
Therefore, according to Descartes sign change, since
$f(x)=3x^5-2x^4-2x^2+x-1=0$ has $3$ sign changes, $f(x)$ has $3$ or $1$ positive roots.
$f(-x)=-3x^5-2x^4-2x^2-x-1=0$ has no sign change. Therefore, $f(x)$ has no negative roots.
