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?

The equation $x^3 - x^2 + 2 = 0$ has three solutions. Therefore we can write it as $(x-a)(x-b)(x-c)=0$. Expanding the second expression and equating terms we get:
$a+b+c = 1$; $ab+bc+ac=0$; $abc=-2$
Now substitute $a = 1+i$. We get $b+c = -i$ and $bc=-1+i$. Which is easily solved.
