Question

求多项式的余式设f(x)为实系数多项式,以x-1除之,余数是7,以x-2除之,余数是12,则f(x)除以（x-1)(x-

Answer 1

设f(x)=(x-1)*g(x)+7;

f(x)=(x-2)k(x)+12;

设g(x)=p(x)*(x-2)+r_1;

------>f(x)=(x-1)*[p(x)*(x-2)+r_1]+7=(x-2)*(x-1)*p(x)+(x-1)*r_1+7=(x-2)k(x)+12;

设(x-1)*r_1+7=(x-2)*q(x)+r_2;

------>f(x) =(x-1)*[p(x)*(x-2)+r_1]+7=(x-2)*(x-1)*p(x)+(x-2)*q(x)+r_2

=(x-2)*[(x-1)*p(x)+q(x)]+r_2

=(x-2)k(x)+12;

------>(x-1)*p(x)+q(x)=k(x) AND r_2=12

------>(x-1)*r_1=(x-2)*[k(x)-(x-1)*p(x)]+5

------>(x-2)*r_1+r_1=(x-2)*[k(x)-(x-1)*p(x)]+5

------>r_1=(x-2)*[k(x)-(x-1)*p(x)-r_1]+5;

因为5 显然不能被x-2整除,故(x-2)除r_1的余式是5；

故(x-2)(x-1)除(x-1)*r_1的余式是5；

设(x-1)*r_1=(x-2)(x-1)*u(x)+5;

so

------>f(x)=(x-1)*[p(x)*(x-2)+r_1]+7=(x-2)*(x-1)*p(x)+(x-1)*r_1+7

=(x-2)*(x-1)*p(x)+(x-2)(x-1)*u(x)+5+7;

=(x-2)*(x-1)*[p(x)+u(x)]+12;

so

余式为12.