Answer
$4$ positive real zeros
$2$ positive real zeros and $2$ imaginary zeros
$4$ imaginary zeros
Work Step by Step
We are given the function:
$$g(x)=2x^4-8x^3+6x^2-3x+1.$$
First we count the number of sign changes:
$$+,-,+,-,+.$$
The coefficients of $g(x)$ have $4$ sign changes, so $g$ has either $4$ positive real zeros, or $2$ positive real zeros or $0$ positive real zeros.
We determine $g(-x)$:
$$\begin{align*}
g(-x)&=2(-x)^4-8(-x)^3+6(-x)^2-3(-x)+1\\
&=2x^4+8x^3+6x^2+3x+1.
\end{align*}$$
We count the number of sign changes:
$$+,+,+,+,+.$$
The coefficients of $g(-x)$ have no change of sign, therefore there is no negative real zero.
To summarize the results, we use a table:
\[ \begin{array}{|c|c|c|c|}
\hline
\text{Positive}& \text{Negative} & \text{Imaginary} & \text{Total} \\
\text{real zeros} & \text{real zeros} & \text{zeros} & \text{zeros}\\
\hline
4 &0 & 0&4\\
\hline
2 & 0 &2&4\\
\hline
0 & 0 & 4&4\\
\hline
\end{array}\]
