If $x=2+5i$ then the value of $x^3-5x^2+$$33x-19$  is equal to

$x=2+5i\Rightarrow x-2=5i$

$\Rightarrow (x-2{)}^2=(5i{)}^2\Rightarrow x^2-4x+4=-25$

$\Rightarrow x^2-4x+29=0$ We now divide 

$x^3-5x^2+33x-19$ by $x^2-4x+29$

$$\begin{gathered} x-1 \\ {x}^2-4x+29\overline{)x^3-5x^2+33x-19} \end{gathered}$$

                          $$\begin{gathered} x^3-4x^2+29x \\ - \oplus - \\ \_\_\_\_\_\_\_\_\_\_\_ \\ \begin{array}{r}-x^2+4x-19\\ -x^2+4x-29\\ - \oplus -\\ \end{array} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \underline{10} \end{gathered}$$

Thus $\begin{array}{l}{x}^3-5x^2+33x-19=(x-1)\left(x^2-4x+29\right)+10=(x-1)\left(0\right)+10=10\\ \end{array}$