## Algebra 2 (1st Edition)

In the given polynomial of degree $5$, there are $5$ zeros. Here, we have $h(x)=x^5-3x^3+8x-10$ By Descartes' Rule, we have $3$ sign changes, thus we have either $3$ or $1$ positive real roots. Now, $h(-x)=-x^5+3x^3-8x+10$ The above polynomial shows that there are $2$ sign changes. By Descartes' Rule, we have $2$ or $0$ negative real roots. Thus, the other two roots must be imaginary. Positive real zeros = 1 or 3 Negative real zeros = 0 or 2 Imaginary zeros = 4, 2 or 0