I seem to be forgetting my college algebra. Can someone help me understand how to find the complex roots of the polynomial: $x^3+2x+2=0$
I tried synthetic devision for the possible real roots of $\pm1,\pm2$ but, alas, it doesn't reduce to a lower degree.
Apparently, the roots are: $x\approx -0.8,0.4\pm1.6i$
What are ways I can find the complex roots of this equation?
Let $x=u+v$.
Hence, $$u^3+v^3+3uv(u+v)+2(u+v)+2=0.$$ Now, let $u^3+v^3+2=0$.
Thus, since $u+v\neq0$, we obtain $$3uv+2=0$$ or $$u^3v^3=-\frac{8}{27},$$ which says $u^3$ and $v^3$ they are roots of the following equation $$t^2+2t-\frac{8}{27}=0.$$
$t_1=-1+\sqrt{\frac{35}{27}}$, $t_2=-1-\sqrt{\frac{35}{27}}$, which gives an unique real root $$\sqrt[3]{-1+\sqrt{\frac{35}{27}}}+\sqrt[3]{-1-\sqrt{\frac{35}{27}}}.$$
By the Viete's theorem we can get two complex roots: $$x_2=-\frac{u+v}{2}+\frac{(u+v)\sqrt3}{2}i$$ and $$x_3=-\frac{u+v}{2}-\frac{(u+v)\sqrt3}{2}i.$$