Question

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

Answer 1

设x^3+kx^2+x+1
=
(x-1)f(x)
f(x)为x表达式
x=1时，式子两边值为0
所以
1+k+1+1=0
k=-3
x^3-3x^2+x+1
=x^3-x^2-2x^2+2x-x+1
=x^2(x-1)-2x(x-1)-1
=(x^2-2x-1)(x-1)
如果要求实数范围内分解因式的话
=(x-1-√2)(x+1+√2)(x-1)

Answer 2

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)

Answer 3

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)

Answer 4

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)

Answer 5

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)