Answer
The equation has one rational root: $~~x = -1$
The equation has either two or four irrational roots.
Work Step by Step
$P(x) = x^5-x^4-x^3-5x^2-12x-6 = 0$
We can list the possible rational zeros:
$\frac{factors~of~6}{factors~of~1} = \pm 1, \pm 2, \pm 3, \pm 6$
Using synthetic division to check each possible rational zero, we find that only $x = -1$ is a rational zero.
We can consider Descartes Rule of Signs.
$P(x)$ has one variation in sign, so it has one positive real zero.
$P(-x) = -x^5-x^4+x^3-5x^2+12x-6 = 0$
$P(-x)$ has four variations in sign, so $P(x)$ has four, two, or zero negative real zeros.
We found above that $x = -1$ is a real zero, so $P(x)$ has four or two negative real zeros.
Thus. $P(x)$ has one positive real zero and two or four negative real zeros.
The only rational zero is $x = -1$, which is a negative real zero.
Therefore, $P(x)$ has one positive irrational zero and one or three negative irrational zeros.
$P(x)$ has either two or four irrational roots.