How to factorized this 4th degree polynomial? I need your help to this polynomial's factorization.
Factorize this polynomials which doesn't have roots in Q. 
$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $ 
P.S.) Are there any generalized method finding 4th degree polynomials factor?

Yes, there is a general method for determining whether or not a quartic is irreducible in $\mathbb Q[x]$:


*

*Does it have rational roots? In your case, the answer is negative, by the rational root theorem.

*Consider the polynomial $q(x)=p\left(x-\frac12\right)=x^4-\frac{19 x^2}2+3x-\frac3{16}$.

*Consider the resolvent cubic of $q(x)$:\begin{align}r(x)&=x^3+2\times\left(-\frac{19}2\right)x^2+\left(\left(-\frac{19}2\right)^2-4\times\left(-\frac3{16}\right)\right)x-3^2\\&=x^3-19x^2+91x-9.\end{align}

*Does it have a rational root which is a perfect square in $\mathbb Q$? Yes: $9(=3^2)$ is such a root.

*Then $q(x)=\left(x^2+3x+\beta\right)\left(x^2-3x+\gamma\right)$, with$$\beta=\frac12\left(-\frac{19}2+3^2-\frac33\right)=-\frac34\text{ and }\gamma=\frac12\left(-\frac{19}2+3^2+\frac33\right)=\frac14.$$

*Since $p(x)=q\left(x+\frac12\right)$, you get a decomposition for $p(x)$:$$p(x)=(x^2+4x+1)(x^2-2x-1).$$
