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

Kassandra Mccall

Kassandra Mccall

Answered question

2022-09-07

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

Answer & Explanation

Xavier Jennings

Xavier Jennings

Beginner2022-09-08Added 9 answers

Fun fact:
x 4 + 1 = ( x 2 + 2 x + 1 ) ( x 2 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 + 2 x 2 + 1 ) ( x 4 2 x 2 + 1 ) .
We can go further. For example, set
x 4 + 2 x 2 + 1 = ( x 2 + a x + b ) ( x 2 a x + 1 / b ) ,
yielding
{ a 2 + b + 1 / b = 2 a / b a b = 0
Rule out a = 0 to obtain b = ± 1 from the second equation, then plug those candidates into the first equation yielding a = ± 2 2 = 2 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
x 8 + 1 = ( x 2 + 2 2 x + 1 ) × ( x 2 2 2 x + 1 ) × ( x 2 + 2 + 2 x + 1 ) × ( x 2 2 + 2 x + 1 ) .
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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