I'm trying to find the roots of P(x) :$$x^3-3x\enspace-4=0$$
First I tried $P(\pm1),\enspace P(\pm2), \enspace P(\pm4)$ , and I found no rational solutions.
So I used the formula for the solutions of a third degree equation:
given $$x^3+px+q=0$$
you have $$x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}\enspace+\enspace \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$
with $p = -3,\enspace q = -4\enspace$ I found the real solution $$x = \sqrt[3]{2+\sqrt{3}}\enspace +\enspace\sqrt[3]{2 - \sqrt{3}}$$
Now how do I find the two complex solutions?
Should I divide $\enspace x^3-3x\enspace-4\enspace$ by $\Bigl(x - (\sqrt[3]{2+\sqrt{3}}\enspace +\enspace\sqrt[3]{2 - \sqrt{3}})\Bigr) $?
Let $\omega=\cos\left(\frac{2\pi}3\right)+\sin\left(\frac{2\pi}3\right)i=-\frac12+\frac{\sqrt3}2i$. Then the roots of the equation are$$\sqrt[3]{2+\sqrt3}+\sqrt[3]{2-\sqrt3},\ \sqrt[3]{2+\sqrt3}\,\omega+\sqrt[3]{2-\sqrt3}\,\omega^2\text{ and }\sqrt[3]{2+\sqrt3}\,\omega^2+\sqrt[3]{2-\sqrt3}\,\omega.$$You will find an explanation here.