将“χ的16次方+x的8次方+1“因式分解
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/15 19:01:54
将“χ的16次方+x的8次方+1“因式分解
![将“χ的16次方+x的8次方+1“因式分解](/uploads/image/z/7145635-67-5.jpg?t=%E5%B0%86%E2%80%9C%CF%87%E7%9A%8416%E6%AC%A1%E6%96%B9%2Bx%E7%9A%848%E6%AC%A1%E6%96%B9%2B1%E2%80%9C%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3)
x^16+x^8+1
=(x^8)^2+2x^8+1-x^8
=(x^8+1)^2-(x^4)^2
=(x^8+1+x^4)(x^8+1-x^4)
=[(x^4)^2+2x^4+1-x^4](x^8-x^4+1)
=[(x^4+1)^2-(x^2)^2](x^8-x^4+1)
=(x^4+1+x^2)(x^4+1-x^2)(x^8-x^4+1)
=[(x^2)^2+2x^2+1-x^2)(x^4-x^2+1)(x^8-x^4+1)
=[(x^2+1)^2-x^2)](x^4-x^2+1)(x^8-x^4+1)
=(x^2+x+1)(x^2-x+1)(x^4-x^2+1)(x^8-x^4+1)
=(x^8)^2+2x^8+1-x^8
=(x^8+1)^2-(x^4)^2
=(x^8+1+x^4)(x^8+1-x^4)
=[(x^4)^2+2x^4+1-x^4](x^8-x^4+1)
=[(x^4+1)^2-(x^2)^2](x^8-x^4+1)
=(x^4+1+x^2)(x^4+1-x^2)(x^8-x^4+1)
=[(x^2)^2+2x^2+1-x^2)(x^4-x^2+1)(x^8-x^4+1)
=[(x^2+1)^2-x^2)](x^4-x^2+1)(x^8-x^4+1)
=(x^2+x+1)(x^2-x+1)(x^4-x^2+1)(x^8-x^4+1)