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?
 A: As the coefficients are real thus the complex roots would appear in conjugate pair. Thus the second root would be $1-i$ and the third root (that is $-1$) can be calculated by using the fact that the sum of the roots of a cubic equation $ax^3+bx^2+cx+d$ is $-b/a$.
A: 
$x^3-x^2+2= (x-1-i)\,x^2 + 2 + i\,x^2\;$ 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":
$$
x^2 + ix - 1 + i = (x^2-1) + i(x+1)=(x+1)(x-1)+i(x+1)=(x+1)(x-1+i)
$$
A: One thing to note is that complex roots of polynomials with real-valued coefficients ALWAYS come in pairs of complex conjugates (in other words, if $a+bi$ is a root, so is $a-bi$). In other words, since you know that $1+i$ is a root, this means you also know that $1-i$ is a root. If you are not aware of this fact, try to prove it for yourself or try to see why! If you would like, let me know and I can provide further justification.
Now that you know $1+i$ and $1-i$ are both roots, you can now see that $x-1-i$ and $x-1+i$ are both linear factors!
$$(x-1-i)(x-1+i)=x^2-2x+2$$
Now, doing the same long division you provide above, you should see that the final linear term is $x+1$!
A: 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.
A: One of the other roots is $1-i,$ since complex roots of real polynomials come in complex conjugate pairs.
The sum of the roots is $-1$ times the coefficient of $x^2.$ So, if $r$ is the third root, $(1+i)+(1-i)+r=1,$ or $r=-1.$
A: 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 $x_3=-1.$
