Answer
$C$ ; 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 $C$.