Question

关于多项式的除法,有关余式的情况

Answer 1

设(7x-6)/(x^2-3*x+2)=a/(x-1)+b/(x-2)

a/(x-1)+b/(x-2)

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

=[(a+b)x-(2a+b)]/(x^2-3x+2)

=(7x-6)/(x^2-3x+2)

所以a+b=7

2a+b=6

a=-1,b=8

(7x-6)/(x^2-3x+2)=8/(x+2)-1/(x-1)