In C[x] one of the roots of the polynomial $V(x)$ is $X1=-1+i$. Find the remaining roots of $V(x)$.
I'm so sorry for the picture, but I need extra help with it. Hope you understand me.
Probably my attempt is not true, can someone help me how to solve it?
Best I can decipher, you are asked for the complex roots of the polynomial $$V(x)=x^6+2x^5+2x^4+3x^2+6x+6.$$
You are given that one of the roots is $x_1=-1+i.$
Since coefficients are real, you correctly deduce that the complex conjugate $x_2=-1-i$ is also a root.
$(x - (-1+i))(x-(-1-i))=x^2+2x+2$ therefore divides $V(x),$ and the quotient is $x^4+3$; that can be seen by inspection, though you did long division. So we are left to find the roots of $x^4+3$.
If $x^4=-3$ then $x^2= \pm \sqrt3i.$ Solutions of $x^2=\sqrt3i$ are $$x_{3,4}=\pm\sqrt[4]3\frac{1+i}{\sqrt2}$$ (it doesn't make sense to write $\sqrt i$) and solutions of $x^2=-\sqrt3i$ are $$x_{5,6}=\pm\sqrt[4]3\frac{1-i}{\sqrt2}.$$