x^3+kx^2-x+3因式分解后有一个因式是x-1,求k的值,将此多项式因式分解
问题描述：
x^3+kx^2-x+3因式分解后有一个因式是x-1,求k的值,将此多项式因式分解
答
x^3+kx^2-x+3因式分解后有一个因式是x-1
将x=1代入x^3+kx^2-x+3=0
1+k-1+3=0
k=-3
x^3-3x^2-x+3
==(x-1)(x^2-2x-3)
=(x-1)(x-3)(x+1)
答
令x＾3+kx＾2-x+3=0
则x＝1是其的一个解
得1+k-1+3=0
k=-3
有x＾3-3x＾2-x+3
则得x＾2(x-3)-(x-3)
(x＾2-1)(x-3)
(x+1)(x-1)(x-3)
答
设f(x)=x^3+kx^2-x+3,则f(1)=0,所以1+k-1+3=0,所以k=-3
因此x^3-3x^2-x+3=x^3-x^2-2x^2+2x-3x-3=(x-1)(x^2-2x-3)=(x-1)(x-3)(x+1)
答
将x=1代入x^3+kx^2-x+3=0
1+k-1+3=0
k=-3
x^3+kx^2-x+3
= x^3-3x^2-x+3
= (x^3-3x^2)-(x-3)
= x^2(x-3)-(x-3)
= (x-3)(x^2-1)
= (x-3)(x+1)(x-1)