Factor $x^6 +5x^3 +8$ I wanted to know, how can I factor $x^6 +5x^3 +8$, I have no idea. Is there any method to know if a polynomial is factored. Just some advice will do.
Help appreciated.
Thanks.
 A: Let $y = x^3$ to obtain $y^2 + 5y + 8 = 0$. This factors as $y = \frac{-5 \pm \sqrt{-7}}{2} = \frac{-5}{2} \pm \frac{\sqrt{7}}{2}i$. These have modulus $r = \sqrt{25/4 + 7/4} = 2\sqrt{2}$. Now solve $5/2 = 2\sqrt{2} \cos \theta$ to find the angle $\theta = \cos^{-1}(5/4\sqrt{2})$. Our two roots correspond to $re^{\pi-\theta}$ and $re^{\pi+\theta}$.
Now we have $x^3 = re^{\pi-\theta}$ and $x^3 = re^{\pi+\theta}$ to contend with. For the former, one root is $x_1 = \sqrt[3]{r}e^{(\pi-\theta)/3}$, so the other two are $x_2 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 2\pi/3)$ and $x_3 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.
Similarly, the roots of the other equation are $x_4 = \sqrt[3]{r}e^{(\pi+\theta)/3}$, $x_5 = \sqrt[3]{r}\exp(\frac{\pi+\theta}{3} + 2\pi/3)$, and $x_6 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.
A: I think here for factoring the polynomial $x^6+5x^3+8$ we must go ahead by the same method of Eric Tressler. But, when we get the roots, it is seen that two complex roots are conjugate and so form the factor $x^2-x+2$. Hence, the factorization of Veritas is obtained. I think with this way Veritas get the factorization.
