x的三次方减去8x的二次方加上19x减去12等于0，求解过程，急急!!!!!!!!!!!!!!!

Question

如题所述

Answer 1

x^3-8x^2+19x-12=0
(x^3-x)-(8x^2-20x+12)=0
x(x^2-1)-4(2x^2-5x+3)=0
x(x+1)(x-1)-4(x-1)(2x-3)=0
(x-1)(x^2+x-8x+12)=0
(x-1)(x-3)(x-4)=0
x=1,x=3,x=4本回答被提问者采纳

Answer 2

x^3-8x^2+19x-12=0
x(x^2-1)-4(2x^2-5x+3)=0
（x-1)(x^2+x-8x+12)=0
(x-1)(x-3)(x-4)=0
x1=1,x2=3,x3=4

Answer 3

分解因式(x-1)(x-3)(x-4)=0,
x1=1,x2=3,x3=4