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?

For all real $k$ we have:
$$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$
$$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1).$$
Now, we'll choose a value of $k$ such that
$$(2k+9)x^2+2(k+3)x+k^2+1=(ax+b)^2.$$
Easy to see that $k=0$ is valid.
In the general we need to solve the following equation.
$$(k+3)^2-(2k+9)(k^2+1)=0.$$
Id est, $$x^2+2x^3-8x^2-6x-1=(x^2+x+k)^2-x^2-k^2-2kx^2-2kx-8x^2-6x-1=$$
$$=(x^2+x+k)^2-((2k+9)x^2+2(k+3)x+k^2+1)=$$
$$=(x^2+x)^2-(9x^2+6x+1)=(x^2+x)^2-(3x+1)^2=(x^2-2x-1)(x^2+4x+1).$$
