Chapter 5 Polynomials and Polynomial Functions - 5.7 Apply the Fundamental Theorem of Algebra - 5.7 Exercises - Skill Practice - Page 384: 35

Positive real root =1 Negative real root= 0 Imaginary roots: 2

In the given polynomial of degree $3$, there are $3$ zeros. Here, we have $g(x)=-x^3+5x^2+12$ By Descartes' Rule, we have $1$ positive real root. Now, $g(x)=-(-x)^3+5(-x)^2+12=x^3+5x+12$ The above polynomial shows that there is no sign change. By Descartes' Rule, we have no negative real root. Thus, the other two roots must be imaginary. Hence, positive real root =1 Negative real root= 0 Imaginary roots: 2