How to factor $-x^3+x^2-2$ Is there an easy way to factor $$f(x)=-x^3+x^2-2\;?$$ 
I have checked step-by-step calculators that all use theorems I am not very familiar with. 
It doesn't seem like a sum or difference of cubes, and grouping is not an option. I tried factoring out the $-x^2$ $$f(x)=-x^2(x-1)-2$$ but that didn't get me anywhere. I am trying to find the roots to graph this function. How do I approach this?
 A: $$-x^3+x^2-2=-1-x^3+x^2-1$$
$$=-(1+x^3)+(x^2-1)$$
$$=-(1+x)(1-x+x^2)+(x-1)(x+1)$$
$$=(x+1)(x-1-(1-x+x^2))$$
$$=(x+1)(-x^2+2x-2)$$
Edit:  At the fourth step, there is a factor of $x+1$ in both terms.  So I factor this out of the expression as follows:
$$-(1+x)(1-x+x^2)+(x-1)(x+1)=(x+1)[-(1-x+x^2)]+(x+1)(x-1)$$
$$=(x+1)[-(1-x+x^2)+(x-1)]$$
$$=(x+1)(-1+x-x^2+x-1)$$
$$=(x+1)(-x^2+2x-2)$$
A: If you can find a root, $r$ then it will factor to $(x-r)(ax^2 + bx + c)$ and then we can you complete the square or use quadratic formula to see if $ax^2 + bx + c$ has rational roots.  If it has roots $q_1$ and $q_2$ this will factor to $a(x-r)(x-q_1)(x-q_2)$
So can we find any roots.  By rational root theorem, any rational roots of $-x^3+x^2 - 2=0$ will be of the form $\pm 1, \pm 2$.  
$x=-1$ is a root as $-(-1)^3 + (-1)^2 -2 = 0$ but $x=1$ is not. $x=2$ nor is $x = -2$.
Dividing by $(x- (-1))$ or $x+1$ we get $\frac {-x^3 +x^2 -2}{x + 1} = \frac {-x^3}{x+1} + \frac {x^2-2}{x+1}=$
$\frac {-x^3- x^2}{x+1} + \frac {x^2 + x^2-2}{x+1}=$
$-x^2 + \frac {2x^2 - 2}{x+1} = -x^2 +\frac {2x^2}{x+1} +\frac {-2}{x+1}=$
$-x^2 + \frac {2x^2 + 2x}{x+1} + \frac {-2x - 2}{x+1}=$
$-x^2 + 2x -2$
So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.  
Can we factor $-x^3 + x^2 -2$?  We could use quadratic formula or complete the square but we do know from above that $x = -1$ is the possible rational solution. And it doesn't work. ($-(-1)^2 + 2(-1)-2 = -3 \ne 0$.)
So $-x^2 + 2x -2$ can not factor.  (By quadratic equation the roots are $\frac {-2\pm \sqrt{4 -8}}{-2}$ and those aren't possible.  But we didn't need to do that.)
So $-x^3 +x^2 -2 = (x+1)(-x^2 + 2x -2)$.  We can, for style points, bring the negative term out to get $-x^3 + x^2 -2 = -(x+1)(x^2 - 2x +2)$.  But we can't factor any further.
A: Every polynomial of degree $3$, with real coefficients, has at least one real root $r$.
So, if you find such $r$, you can write:
$$
P(x)=-x^3+x^2-2=(x-r)\cdot p_2(x)\tag{1}
$$
where $p_2(x)$ is a second degree polynomial in $x$. This was also discussed in the answer by fleablood and you can do this before some further factoring, if necessary.
How to find a root $r$ of $P(x)$? It must be in the form $r=\dfrac{p}{q}$, with $p$ dividing $-2$ (which is the coefficient of $x^0$) and $q$ dividing $-1$ (which is the coefficient of the highest degree monomial). Easily you find that such values are those in the set
$$
\{\pm 1, \pm 2\}.
$$
If you calculate $P(1)$, $P(-1)$, $P(2)$ and $P(-2)$ you can check that only $P(-1)$ is $0$, hence, from (1):
$$
-x^3+x^2-2=(x+1)\cdot p_2(x).
$$
You have several methods to divide polynomials, and you can find out that $p_2(x)=-x^2+2x-2$: this one has $\Delta < 0$, meaning that it cannot be further reduced.
It follows that:
$$
-x^3+x^2-2 = (x+1) (-x^2+2x-2).
$$
