## Algebra 2 (1st Edition)

In the given polynomial of degree $5$ , there are $5$ zeros. Here, we have $h(x)=x^5-2x^3-x^2+6x+5$ By Descartes' Rule, we have two sign changes, thus we have either $2$ or $0$ positive real roots. Now, $h(-x)=-x^5+2x^3-x^2-6x+5$ The above polynomial shows that there are $3$ sign changes. By Descartes' Rule, we have $3$ or $1$ negative real roots. Thus, the other two roots must be imaginary. Hence, positive real roots = 0 or 2 Negative real roots = 1 or 3 Imaginary roots = 4, 2 or, 0