二元三次方程

2x^3+4x-6
=2x^3-2x^2+2x^2-2x+6x-6
=(2x^3-2x^2)+(2x^2-2x)+(6x-6)
=2x^2(x-1)+2x(x-1)+6(x-1)
=(x-1)(2x^2+2x+6)

这道题是因式分解,不是解方程!!!

因各项系数之和2+4-6=0;方程2x^3+4x-6=0,明显有解x=1,所以有x-1因式;2x^3+4x-6=(2x^3-2x^2)+(2x^2-2x)+(6x-6)=2x^2(x-1)+2x(x-1)+6(x-1)=(x-1)(2x^2+2x+6)=2(x-1)(x^2+x+3);