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?

Maybe all has been said already.
Polynomial with real coefficients:
1)Complex roots occur in pairs, one is the complex conjugate of the other.
(Complex root theorem)
$x_1=1+i,$ $x_2=1-i;$
2)By inspection the real root is $x_3=-1.$
