Question

已知关于x的多项式x^3+kx^2－x+3因式分解后有一个因式是x－1

已知关于x的多项式x^3+kx^2－x+3因式分解后有一个因式是x－1
（1）求实数k的值；
（2）将此多项式因式分解

Answer 1

设另一因式是A
x³+kx²-x+3=A(x-1)
则x=1时，x-1=0
所以右边是0
所以左边也是0
所以x=1
1+k-1+3=0
所以k=-3

原式=x³-x²-(2x²+x-3)
=x²(x-1)-(2x+3)(x-1)
=(x-1)(x²-2x-3)
=(x-1)(x+1)(x-3)

Answer 2

f(x) =x^3+kx^2-x+3
f(1) = 1+k-1+3 =0
k=-3
f(x) =x^3-3x^2-x+3
f(x) = (x-1) (x^2+ax -3)
coef. of x
-a-3 = -1
a= -2
f(x) = (x-1) (x^2-2x -3)
=(x-1)(x-3)(x+1)

Answer 3

(x^2+ax+b)(x-1)
=x^3-x^2+ax^2-ax+bx-b
=x^3+(a-1)x^2-(a-b)x-b
b=-3
a-b=1
a=-2
k=a-1=-3
(x^2-2x-3)(x-1)

=(x+1)(x-3)(x-1)