Question

已知多项式x^3+2x+m（m是常数）有一个因式是x-1,求m的值及分解这个多项式.

Answer 1

设x^3+2x+m=(x-1)(x^2+ax+b)

=x^3+ax^2+bx-x^2-ax-b

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

a-1=0 b-a=2 -b=m

a=1 b=3 m=-3

x^3+2x-3=(x-1)(x^2+x+3)