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? 


One factor is $$x^2+2x+2=0$$ which can be solved by the quadratic formula. Solving $$x^4+3=0$$ we get
$$\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1-i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1-i)$$
Substituting $$x^2=t$$ then we get $$t^2+(\sqrt{3})^2=0$$ By the formula $$a^2+b^2=(a+bi)(a-bi)$$ you will get the solutions.
