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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.

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    $\begingroup$ Let $y = x^3$ and use the quadratic formula. $\endgroup$ Jun 29, 2013 at 9:25
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    $\begingroup$ $x^6+5x^3+8= (x^2-x+2)*(x^4+x^3-x^2+2x+4)$ $\endgroup$
    – Veritas
    Jun 29, 2013 at 9:47
  • $\begingroup$ @Veritas, how did you get that? $\endgroup$
    – lhf
    Jun 29, 2013 at 11:27
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    $\begingroup$ @lhf the equation becomes (x^2)^3 + (-x)^3 + (2)^3 -3*(x^2)*(-x)(2). which is a^3 + b^3 + c^3 -3abc. $\endgroup$
    – Shobhit
    Jun 29, 2013 at 11:33

2 Answers 2

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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 $x_6 = \sqrt[3]{r}\exp(\frac{\pi-\theta}{3} + 4\pi/3)$.

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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.

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