求函数y=x^3-2x^2-x+2的零点

来源：百度知道编辑：UC知道时间：2024/06/01 17:14:16

请告诉我解题步骤，谢谢啦

因式分解
y=x^2(x-2)-(x-2)
=(x-2)(x^2-1)
=(x-2)(x-1)(x+1)
所以零点是2,1,-1

y=x^3-2x^2-x+2
=x^3-x^2-x^2-x+2
=x^2(x-1)-(x^2+x-2)
=x^2(x-1)-(x+2)(x-1)
=(x-1)(x^2-x-2)
=(x-1)(x+1)(x-2)
y=x^3-2x^2-x+2的零点:x=1,x=-1,x=2

y=x^3-2x^2-x+2=x^2(x-2)-(x-2)=(x^2-1)(x-2)=(x-1)(x+1)(x-2)
所以函数y=x^3-2x^2-x+2的零点为x=1,x=-1,x=2

原式=x^2(x-2)-(x-2)
=(x^2-1)(x-2)=(x+1)(x-1)(x-2)
所以零点为x=-1,1和2