Question

如何求解一元三次方程3x^3-8x^2+6x-1=0 急求 麻烦能帮忙解答一下 请写出具体过程 谢谢

Answer 1

3x^3-8x^2+6x-1=3x^3-3x^2-5x^2+5x+x-1=3x^2(x-1)-5x(x-1)+(x-1)

=(x-1)(3x^2-5x+1)=0

x-1=0,x=1

3x^2-5x+1=0,判别式=25-4*3*1<0,无解

3x^3-8x^2+6x-1=0方程的解为x=1

Answer 2

因式分解
3x³-3-8x²+6x+2=0
3(x³-1)-2(4x²-3x-1)=0
3(x-1)(x²+x+1)-2(4x+1)(x-1)=0
(x-1)(3x²+3x+3-8x-2)=0
(x-1)(3x²-5x+1)=0
x-1=0,3x²-5x+1=0
x=1,x=(5-√13)/2,x=(5+√13)/2