1. For the polynomial in $x$
$$
x^{3}-g x^{2}+(g+m) x+6 m^{2}+6 m-8
$$

to be divisible by $x-2$, and when $m$ is a certain real number, $g$ has a minimum value. Then, this polynomial should be $\qquad$ .
$=, x^{3}-\frac{4}{3} x^{2}+\frac{2}{3} x-\frac{28}{3}$.
The remainder when the given polynomial is divided by $x-2$ is
$$
2(4+m-g)+\left(6 m^{2}+6 m-8\right) \text {. }
$$

Since it can be divided exactly, the above expression is complex, i.e., $g=3 m^{2}+4 m$. When $m$ $=-\frac{2}{3}$, $g_{\text {min }}=\frac{4}{3}$, at this time $g+m=\frac{2}{3}, 6 m^{2}+6 m-8$ $=-\frac{28}{3}$.
Therefore, this polynomial is $x^{3}-\frac{4}{3} x^{2}+\frac{2}{3} x-\frac{28}{3}$.