Note that $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$; this factorization was guesswork on my part, based on the observation that $(x^2 - \mu_1)(x^2 - \mu_2) = x^4 - (\mu_1 + \mu_2)x^2 + \mu_1 \mu_2$; so I looked for integers $\mu_1$, $\mu_2$ such that $\mu_1 + \mu_2 = 1$ and $\mu_1 \mu_2 = -12$; it was easy to make such a guess. But this factorization could have been done systematically by setting $y = x^2$, as several others have suggested, and then realizing the roots of the quadratic $y^2 - y -12 = 0$ are $4, -3$; this of course can be done with aid of the quadratic formula. Once we have $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$, we can factor further by using $x^2 - 4 = (x + 2)(x - 2)$; thus $x^4 - x^2 -12 = (x + 2)(x - 2)(x^2 + 3)$;
we can't go further over the reals since $x^2 + 3$ has no real zeroes.
That's how I'd do it, in any event.
Hope this helps. Cheerio,
and as always,