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?

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 $-\sqrt{2}-i$
