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

Question

So the problem is :

$x^4-4x^3-x^2-8x+4=0$, find all solutions

A tip that I have gotten, is to divide both sides by $x^2$. I've tried so, but I do not manage to see any further. Do anyone know how this tip could help me?

(Yes, I'm aware that the polynomial above can be factorized into two degree 2 polynomials, which promptly gives me the answer. But that factorization would be extremely hard to spot, which is why I'm asking about the dividing)

Thanks in advance :)

Edit: meant to write $x^2$, not $2$

Answer 1

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 $(x^2-5x+2)(x^2+x+2)$.

Answer 2

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

Answer 3

$ 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.