Solve this $6$-th degree polynomial equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$ The question is as follows: 

Solve the equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$, one root being $\sqrt{2}+i$

It's trivial that another root is $\sqrt{2}-i$. But, I can go no further. Can anyone please help me how to solve this?
 A: In addition to $\sqrt{2} -i$, we have that $-\sqrt{2}+i$ and $-\sqrt{2}-i$ are also guaranteed to be roots.  (These are the remaining Galois conjugates, which always take the specified forms in the case of a sum of square roots like $\sqrt{2}+i$.)
To determine the other roots, divide your given polynomial $f(x)$ by $$\begin{align}(x-\sqrt{2}-i)(x-\sqrt{2}+i)(x+\sqrt{2}-i)(x+\sqrt{2}+i)
&= (x^2-2\sqrt{2}x+3)(x^2+2\sqrt{2}x+3) \\
 &= x^4-2x^2+9 \end{align}$$
This yields $f(x) = (x^2-x-6)(x^4-2x^2+9)$, so the remaining roots are the roots of $x^2-x-6 = (x-3)(x+2)$, namely -2 and 3.
A: As @avs answered, after dividing  the polynomial by $$(x - \sqrt{2} + i)(x - \sqrt{2} - i)$$ we obtain $$x^4+\left(2 \sqrt{2}-1\right) x^3-\left(3+2 \sqrt{2}\right) x^2-\left(3+12
   \sqrt{2}\right) x-18=0$$ By inspection $x=-2$ and $x=3$ are roots. Continuing the division, we are left with $$x^2+2 \sqrt{2} x+3=0$$ the roots of which being $-\sqrt{2}+i$ and $-\sqrt{2}-i$
A: Divide the polynomial by $(x - \sqrt{2} + i)(x - \sqrt{2} - i)$.  The resulting polynomial is of degree 4, so there is a formula for finding its roots: https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots
