如果(x-1)(x²+ax+b)=x³-6x²+11x-6求常数a,b的值急

如果(x-1)(x²+ax+b)=x³-6x²+11x-6求常数a,b的值急
如果(x-1)(x²+ax+b)=x³-6x²+11x-6求常数a,b的值
急

如果(x-1)(x²+ax+b)=x³-6x²+11x-6求常数a,b的值急
x³-6x²+11x-6
=(x³-x²)-(5x²-11x+6)
=x²(x-1)-(x-1)(5x-6)
=(x-1)(x²-5x+6)
=(x-1)(x²+ax+b)
∴ a=-5,b=6