Finding all roots of $x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$, knowing that one root is $x=-1+i$. 
Find all roots of
  $$V(x) = x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$$
  knowing that one root is $x=-1+i$.

Sorry for the picture. I found two roots of the polynomial and also found an equation which can help me to find all of the rest. Can someone help me what do I need to do next? 


To solve $$x^2+2x+2=0$$ use $$x^2+2x+1=-1$$ hence $$(x+1)^2=-1$$ which gives the solutions $\ -1-i\ $ and $\ -1+i\ $
To solve $$x^4+3=0$$ first note that the absolute value of the solutions must be $3^{\frac{1}{4}}$ and then use the $4$ fourth roots of $-1$ , being $$\pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2}i$$ which , multiplied with $3^{\frac{1}{4}}$, give the $4$ other solutions.
