已知多項式f(x)除以(x-1)和(x-2)所得的余數(shù)分別為1和2,
試求f(x)除已(x-1)(x-2)所得的余式.
f(x)=g(x)(x-1)+1.(1)
f(x)=h(x)(x-2)+2.(2)
(1)*(x-2):--->(x-2)f(x)=g(x)(x-1)(x-2)+(x-2).(3)
(2)*(x-1):--->(x-1)f(x)=h(x)(x-2)(x-1)+(2x-2).(4)
即：f(x)除以(x-1)(x-2)所得的余式 為 x