jmc

Since $x^2-3x+2$ factors as $(x-1)(x-2)$ we can divide $x^2-3x+2$ in two steps using synthetic division as follows. First we divide by $x-1$. $$\begin{array}{rrrrrr}\text{multicolumn}1r|1& 1& -3& 4& 11& -9\\ \text{multicolumn}1r|& & 1& -2& 2& 13\\ \text{cline}2-6& 1& -2& 2& 13& \text{multicolumn}1|r4\end{array}$$Thus, we find that $x^4-3x^3+4x^2+11x-9=(x-1)(x^3-2x^2+2x+13)+4$. Then we divide the quotient $x^3-2x^2+2x+13$ by $x-2$. $$\begin{array}{rrrrr}\text{multicolumn}1r|2& 1& -2& 2& 13\\ \text{multicolumn}1r|& & 2& 0& 4\\ \text{cline}2-5& 1& 0& 2& \text{multicolumn}1|r17\end{array}$$Thus, $x^3-2x^2+2x+13=(x-2)(x^2+2)+17$. Substituting this expression into our first equation gives $$\begin{array}{rl}{x}^4-3x^3+4x^2+11x-9& =(x-1)(x^3-2x^2+2x+13)+4\\ & =(x-1)[(x-2)(x^2+2)+17]+4\\ & =(x-1)(x-2)(x^2+2)+17(x-1)+4\\ & =(x-1)(x-2)(x^2+2)+17x-13.\end{array}$$The quotient is $x^2+2$ and the remainder is $\boxed{17x-13}$.