## Algebra 2 (1st Edition)

In the given polynomial of degree $3$, there are $3$ zeros. Here, we have $g(x)=x^3-4x^2+8x+7$ By Descartes' Rule, we have either $2$ or $0$ positive real roots. Now, $g(x)=-x^3-4x^2-8x+7$ The above polynomial shows that there is one sign change. By Descartes' Rule, we have $1$ negative real root. Thus, the other two roots must be imaginary. Hence, positive real roots = 2 or 0 Negative real root = 1 Imaginary roots = 2 or 0