x^3-3x^2-6x+8=0怎么求?

2024-11-15 18:56:26
推荐回答(5个)
回答1:

分解因式:
x³ - 3x² - 6x + 8
=x³ - x² - 2x² + 2x - 8x + 8
=(x³ - x²) - (2x² - 2x) - (8x - 8)
=x²(x - 1) - 2x(x - 1) - 8(x - 1)
=(x - 1) (x² - 2x - 8)
= (x - 1)(x + 2)(x - 4)

所以
x³ - 3x² - 6x + 8= (x - 1)(x + 2)(x - 4) = 0
所以 x1 = 1
x2 = - 2
x3 = 4

回答2:

回答3:

x^3-3x^2-6x+8=(x-1)*(x-4)*(x+2)=0
所以方程的解有3个,分别是 1, 4和-2

回答4:

回答5:

x³-3x²-6x+8=0
x³-x²-2x²-6x+8=0
(x²(x-1)-2(x²+3x-4)=0
x²(x-1)-2(x-1)(x+4)=0
(x-1)[x²-2(x+4)]=0
(x-1)(x²-2x-4)=0
x-1=0
x1=1
x²-2x-4=0
x²-2x+1-5=0
(x-1)²=5
x-1=±√5
x=1±√5
x2=1+√5
x3=1-√5