Finding the roots $x^4-4x^3-x^2-8x+4=0$ (contest math)

Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(x^2-5x+2)(x^2+x+2)$.

$ x^4 - 4x^3 - x^2 -8 x + 4$ to try to factor it as a product of two polynomials with degree two I will try this

$ x^4 -4x^3 -x^2-8x+4=(x^2+ax+c)(x^2+dx+e) $ but the constant term is 4 so we have two choices $ c=1, e=4$ or $ c=2, e=2$ if you choose the second you get the equations $ a+d=-4, 4+ad=-1, 2a+2d=-8$ if you solve them you come up with a solution or maybe there is not a solution.

Factor as $$(x^2-5x+2)(x^2+x+2)=0$$