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?

Question

I have the polynomial       x    8    +  1, I know that there&#039;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?

Xavier Jennings · Accepted Answer

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.

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?

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