#ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#為整係數,所以b,c亦為整數#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#又f(3/2)=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以(27/8)a+(9/4)b+(3/2)c+a+1=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# 35a+8=6(3b+2c)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#令3b+2c為k,所以35a+8=6k#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可看出a之一般式為2+6m,m=0,1,2,3,...,83,共84個解#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#又3b+2c之係數已互質,故任意整數組合皆可能。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以84個#ed_op#/DIV#ed_cl#GFIF 寫到:若2x-3為整係數多項式f(x)=ax^3+bx^2+cx+(a+1)之因式,其中a為1~500這500個正整數之一,則此種a值有幾個?