Kassandra Mccall

2022-09-07

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

Xavier Jennings

Beginner2022-09-08Added 9 answers

Fun fact:

${x}^{4}+1=({x}^{2}+\sqrt{2}x+1)({x}^{2}-\sqrt{2}x+1).$

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,

${x}^{8}+1=({x}^{4}+\sqrt{2}{x}^{2}+1)({x}^{4}-\sqrt{2}{x}^{2}+1).$

We can go further. For example, set

${x}^{4}+\sqrt{2}{x}^{2}+1=({x}^{2}+ax+b)({x}^{2}-ax+1/b),$

yielding

$\{\begin{array}{ll}-{a}^{2}+b+1/b& =\sqrt{2}\\ a/b-ab& =0\end{array}$

Rule out $a=0$ to obtain $b=\pm 1$ from the second equation, then plug those candidates into the first equation yielding $a=\sqrt{\pm 2-\sqrt{2}}=\sqrt{2-\sqrt{2}}$ 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

$\begin{array}{lll}{x}^{8}+1& =& ({x}^{2}+\sqrt{2-\sqrt{2}}x+1)\times ({x}^{2}-\sqrt{2-\sqrt{2}}x+1)\\ & \times & ({x}^{2}+\sqrt{2+\sqrt{2}}x+1)\times ({x}^{2}-\sqrt{2+\sqrt{2}}x+1).\end{array}$

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.

${x}^{4}+1=({x}^{2}+\sqrt{2}x+1)({x}^{2}-\sqrt{2}x+1).$

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,

${x}^{8}+1=({x}^{4}+\sqrt{2}{x}^{2}+1)({x}^{4}-\sqrt{2}{x}^{2}+1).$

We can go further. For example, set

${x}^{4}+\sqrt{2}{x}^{2}+1=({x}^{2}+ax+b)({x}^{2}-ax+1/b),$

yielding

$\{\begin{array}{ll}-{a}^{2}+b+1/b& =\sqrt{2}\\ a/b-ab& =0\end{array}$

Rule out $a=0$ to obtain $b=\pm 1$ from the second equation, then plug those candidates into the first equation yielding $a=\sqrt{\pm 2-\sqrt{2}}=\sqrt{2-\sqrt{2}}$ 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

$\begin{array}{lll}{x}^{8}+1& =& ({x}^{2}+\sqrt{2-\sqrt{2}}x+1)\times ({x}^{2}-\sqrt{2-\sqrt{2}}x+1)\\ & \times & ({x}^{2}+\sqrt{2+\sqrt{2}}x+1)\times ({x}^{2}-\sqrt{2+\sqrt{2}}x+1).\end{array}$

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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