$C$ ; 2 Positive real zeros, 1 negative real zero, and 2 imaginary zeros.
Work Step by Step
Here, we have $f(x)=x^5-4x^3+6x^2+12x-6$ By Descartes' Rule, we have either $3$ or $1$ positive real zero. Now, $f(-x)=-x^5+4x^3+6x^2-12x-6$ By Descartes' Rule, we have $2$ or $0$ negative real roots. The possible combinations of zeros for $f(x)$ are given as following: A) 2 Positive real zeros and 3 Negative real zeros B) 0 Positive real zeros, 3 Negative real zeros and 2 imaginary zeros C) 2 Positive real zeros, 1 Negative real zero, and 2 imaginary zeros D) 0 Positive real zeros, 1 Negative real zero, and 4 imaginary zeros Hence, our answer is $C$.