[(x³+x+1)(x²-2x+3)+x+1]/(x-1)
={(x³+x+1)[(x-1)^2+2]+x+1}/(x-1)
=(x-1)(x³+x+1)+[2(x³+x+1)+x+1]/(x-1)
=(x-1)(x³+x+1)+(2x³+3x+3)/(x-1)
=(x-1)(x³+x+1)+(2x³-2+3x-3+8)/(x-1)
=(x-1)(x³+x+1)+2(x^2+x+1)+3+8/(x-1)
余式是8