Question

在复数范围内因式分解:2x的平方+x+1请写一下过程

Answer 1

先求方程的根
x=-(-1±√(1-4*1*2))/2*2=(-1±i√7)/4
从而分解为
2x^2+x+1=2(x-(-1+i√7)/4)(x-(-1-i√7)/4)=2(x+(1-i√7)/4)(x+(1+i√7)/4)

Answer 2

解：
2x²+x+1
=2[x²+(1/2)x]+1
=2[x²+2×(1/4)x+(1/4)²]-1/8+1
=2(x+1/4)²+7/8
=2(4x+1)²/16+7/8
=[(4x+1)²+7]/8
=[(4x+1)²-(i√7)²]/8
=(4x+1+i√7)(4x+1-i√7)/8