这一题简单啊.显然
f(x)=a(x+2)+14 则f(-2)=14;
同理f(x)=b(x-3)-6;则f(3)=-6;
下设f(x)=c(x-+2)(x-3)+d;
d至多为一次多项式吧.这样可以知道d(-2)=14;d(3)=-6;
从而d=gx+h满足
-2g+h=14;
3g+h=-6;
解得g=8,h=-30;
从而remainder is 8x-30
有关PolynomialsWhen f(x) is divided by x+2 and x-3,the remaind
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