<s>[INST] 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. 
 [/INST] The way I factor
$x^3 - 2x - 4 \tag 1$
is to notice that $2$ is a root; this I do by intuition and a good amount of experience.  $2$ looks like a good guess to me since the non-leading coefficients are divisible by $2$; once I find that $2$ is a zero of (1), I use synthetic division to find $q(x)$ such that
$x^3 - 2x - 4  = (x - 2)q(x), \tag 2$
thus: $x - 2$ into $x^3$ yields $x^2$; 
$x^3 - 2x - 4 - x^2(x - 2) = 2x^2 - 2x - 4; \tag 3$
$x -2$ in to $2x^2$ yields $2x$:
$2x^2 - 2x - 4 - 2x(x - 2) = 2x - 4; \tag 4$
$x - 2$ into $2x - 4$ yields $2$, and since
$2x - 4 - 2(x - 2) = 0, \tag 5$
we are done.  Gathering the partial quotients yields $x^2 + 2x + 2$, and it is easily checked that
$x^3 -2x -4 = (x - 2)(x^2 + 2x + 2); \tag 6$
now if we want to go further we can use the quadratic formula to find the roots $\mu_\pm$ of $x^2 + 2x + 2$:
$\mu_\pm = \dfrac{-2 \pm \sqrt{-4}}{2} = \dfrac{-2 \pm 2i}{2} = -1 \pm i; \tag 7$
we easily check
$x^2 + 2x + 2 = (x + (1 + i))(x + (1 - i)); \tag 8$
therefore
$x^3 - 2x - 4 = (x - 2)(x + (1 + i))(x + (1 - i)). \tag 9$
And that's the way I factor it!
</s>