X4+1因式分解百度之前查了的说做不出来,google也没搜索到,这个怎么解?

令 x⁴= -1 = -1 + i×0 = cos(2k+1)π + isin(2k+1)π = e^[i(2k+1)π]
∴ x = e^[¼i(2k+1)π]
k=0,x.= e^[¼iπ] = cos(¼π) + isin(¼π) = √2/2 + i√2/2
k=1,x₁= e^[¾iπ] = cos(¾π) + isin(¾π) = -√2/2 + i√2/2
k=2,x₂= e^[5iπ/4] = cos(5π/4) + isin(5π/4) = -√2/2 - i√2/2
k=3,x₃= e^[7iπ/4] = cos(7π/4) + isin(7π/4) = √2/2 - i√2/2
k=4,x₄= e^[9iπ/4] = cos(9π/4) + isin(9π/4) = √2/2 + i√2/2
.
∴ x⁴+ 1 = [x-(√2/2+i√2/2)][x-(-√2/2+i√2/2)][x-(-√2/2-i√2/2)][x-(√2/2-i√2/2)]
= (x-√2/2-i√2/2)(x+√2/2-i√2/2)(x+√2/2+i√2/2)(x-√2/2+i√2/2)

其实一楼和二楼说得都不是很完整
把 X4+1 在 实数范围 分解
是
X4+1
=x4+2x²+1-2x²
=（x²+1）-2x²
=（x²+根号2×x+1）（x²-根号2×x+1）
接下来的 在复数范围 分解
二楼十分正确，但正如一楼所说 “初中用不着”
至于 ...

正常的实数无法分解，用复数可以分解
1=-i^2
X4+1
=(x^2+i)(x^2-i)
后面还是复数分解，初中用不着