jmc

Let $p(x)=x^2-4x+3$. A number $c$ is not in the domain of $g$ if and only if $p(c)=0$. Hence we have, $$c^2-4c+3=0.$$Factoring gives us $$(c-3)(c-1)=0.$$Solving for $c$ gives us $1$ and $3$. Hence the domain of $g$ is $\boxed{(-\infty ,1)\cup (1,3)\cup (3,\infty )}$.