Kassandra Mccall

2022-09-07

I have the polynomial ${x}^{8}+1$, I know that there's no root for solve this in $\mathbb{R}\left[x\right]$ but i want to factorize this to the minimal expression. This is possible or this is irreducible?

Xavier Jennings

Fun fact:
${x}^{4}+1=\left({x}^{2}+\sqrt{2}x+1\right)\left({x}^{2}-\sqrt{2}x+1\right).$
This can be derived by setting ${x}^{4}+1$ equal to a product of two monic quadratics with unknown coefficients and then solving for said coefficients little by little. As a consequence,
${x}^{8}+1=\left({x}^{4}+\sqrt{2}{x}^{2}+1\right)\left({x}^{4}-\sqrt{2}{x}^{2}+1\right).$
We can go further. For example, set
${x}^{4}+\sqrt{2}{x}^{2}+1=\left({x}^{2}+ax+b\right)\left({x}^{2}-ax+1/b\right),$
yielding
$\left\{\begin{array}{ll}-{a}^{2}+b+1/b& =\sqrt{2}\\ a/b-ab& =0\end{array}$
Rule out $a=0$ to obtain $b=±1$ from the second equation, then plug those candidates into the first equation yielding $a=\sqrt{±2-\sqrt{2}}=\sqrt{2-\sqrt{2}}$ with b=1. The second factor of ${x}^{8}+1$ that is written above is similarly reducible. I leave the details as an exercise. The full factorization is
$\begin{array}{lll}{x}^{8}+1& =& \left({x}^{2}+\sqrt{2-\sqrt{2}}x+1\right)×\left({x}^{2}-\sqrt{2-\sqrt{2}}x+1\right)\\ & ×& \left({x}^{2}+\sqrt{2+\sqrt{2}}x+1\right)×\left({x}^{2}-\sqrt{2+\sqrt{2}}x+1\right).\end{array}$
over the real numbers R. With the quadratic formula applied to the above you can get the roots to ${x}^{8}+1$ exactly (they are precisely the primitive 16th roots of unity) in the form of nested radicals, and hence the full factorization in C.
By the way, I should mention that the only nonlinear polynomials over R that are irreducible are quadratics with negative discriminant. This is because the nonreal roots of any real-coefficient polynomial can be paired off into conjugates, and then each conjugate pair of linear factors can be put together to obtain a real quadratic, thus every real-coefficient polynomial can be factored into real linear and quadratic factors.

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