# Given $x=2+5i$, the value of $x^3-5x^2+33x-19$ is

$\begin{array}{1 1}(A)10 \\ (B) 0 \\ (C) -19 \\(D) none\;of\;these \end{array}$

$x=2+5i$
$(x-2)^2=(5i)^2$
$\Rightarrow x^2-4x+29=0$
Dividing $x^3-5x^2+33x-19$ by $x^2-4x+29$ we get,
$(x^3-5x^2+33x-19)=(x-1)(x^2+4x+29)+10$
Putting $x=2+5i$
$(x^3-5x^2+33x-19)=(x-1)(0)+10$
$\Rightarrow 10$
Hence (A) is the correct answer.