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?
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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.
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.