<s>[INST] The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
    x^2 && \hbox{} \\[-3pt]
   x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
      \underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
      2 +i.x^2
  \end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
 [/INST] 
$x^3-x^2+2= (x-1-i)\,x^2 + 2 + i\,x^2\;$ How to pursue by this or some other approach?

The posted answers cover the other approaches.
OP's approach is also salvageable, but first the polynomial division must be taken to completion:
$$
x^3-x^2+2= \left(x-1-i\right)\left(x^2 + ix - 1 + i\right)
$$
The quotient can be further factored either with the quadratic formula, or directly "by inspection":
$$
x^2 + ix - 1 + i = (x^2-1) + i(x+1)=(x+1)(x-1)+i(x+1)=(x+1)(x-1+i)
$$
</s>