jmc

Dividing $x^4+2$ by $x-2,$ we get $$x^4+2=(x-2)(x^3+2x^2+4x+8)+18.$$Dividing $x^3+2x^2+4x+8$ by $x-2,$ we get $$x^3+2x^2+4x+8=(x-2)(x^2+4x+12)+32.$$Thus, $$\begin{array}{rl}{x}^4+2& =(x-2)(x^3+2x^2+4x+8)+18\\ & =(x-2)((x-2)(x^2+4x+12)+32)+18\\ & =(x-2{)}^2(x^2+4x+12)+32(x-2)+18\\ & =(x-2{)}^2(x^2+4x+12)+32x-46,\end{array}$$so the remainder is $\boxed{32x-46}.$