jmc

We can take a factor of $x-2$ out of $(x-2{)}^4-(x-2)=0,$ to get $$(x-2)[(x-2{)}^3-1]=0.$$Then by difference of cubes, $(x-2)-1=x-3$ is also a factor, so $$(x-2)(x-3)[(x-2{)}^2+(x-2)+1]=0.$$This simplifies to $(x-2)(x-3)(x^2-3x+3)=0.$ Hence, $k=\boxed{3}.$