✔ 最佳答案
Sol
(1)
設f(x)=a(2n)x^(2n)+a(2n-1)x^(2n-1)+…+a(1)x+a(0)
a(2n)+a(2n-1)+a(2n-2)+…+a(1)+a(0)=10
a(2n)+a(2n-2)+a(2n-4)+…+a(2)+a(0)=20
a(2n-1)+a(2n-3)+a(2n-5)+…+a(1)=10-20=-10
f(-1)=a(2n)-a(2n-1)+a(2n-2)-a(2n-3)+….+a(2)-a(1)+a(0)=20-(-10)=30
f(x)=Q(x)(x-3)+2
f(-1)=Q(-1)*(-4)+2=30
Q(-1)=-7
Q(x)除以x+1的餘式-7
(2)
f(x)=q(x)(x-2)^2(x+3)^2+a(x-2)^2(x+3)+b(x-2)^2+31x-51
f(-3)=b*(-5)^2-93-51=-4*(-3)-6
25b=150
b=6
6(x--2)^2+31x--51
=6(x^2--4x+4)+31x--51
=6x^2+7x--27