已知多项式3X的3次方-13X的2次方+18x+m能被(x-1)(x-2)整除,其商为3x+n,求m,n的值
问题描述：
已知多项式3X的3次方-13X的2次方+18x+m能被(x-1)(x-2)整除,其商为3x+n,求m,n的值
答
m=-8,n=-4．
提示3x3-13x2+18x+m=(x-1)(x-2)(3x+n)=3x3+(n-9)x2+(6-3n)x+2n．能详细点吗?解整除时有：被除式=除式*商式所以3x3-13x2+18x+m=(x-1)(x-2)(3x+n)3x3-13x2+18x+m=3x3+(n-9)x2+(6-3n)x+2n．所以 -13=n-9 m=2n得m=-8，n=-4．