How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?
To me, this polynomial seems unfactorable. 
But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.
So I got to solve backward:
$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^2-4x=0$$
$$x(x^2+2x+2)-2(x^2+2x+2)=0$$
But I would have thought of this way to factor if I didn't look at the answer key. 

It is better to go with Rational root theorem as suggested. I just want to write another way to see but clearly will not always work out that nicely. 
\begin{equation}
x(x^2-2)-4
\end{equation}
instead of $x^2-2$, write $x^2-4$(because it is $(x-2)(x+2)$) and see if it is possible to factor:
\begin{equation}
x(x^2-4) + 2x  -4 = x(x-2)(x+2) +2(x-2) = (x-2)(x^2+2x+2)
\end{equation}
There is a guess work here, but for example if you start with $x(x^2-2)-1$, you might wanna consider $x^2-1$ instead of $x^2-4$.
