## Algebra 2 (1st Edition)

Published by McDougal Littell

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

#### Answer

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

#### Work Step by Step

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

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