>>417
①③の場合a,bが存在
(a,b)=(1,3),(2,2) x=1
x^3+x^2-5x+3=(x-1)(x^2+2x-3)=(x-1)^2(x+3)
x^3+2x^2-5x+2=(x-1)(x^2+3x-2)