已知f(x)除于x²+x+1和(x+1)²时，餘式分別为x+5及x-1，求f(x)除于(x²+x+1)(x+1)的餘式?

f(x) ÷ (x^2+x+1) 餘式 x+5
f(x) ÷ (x+1)^2 餘式 x-1
也就是說
   f(x) = p(x)(x^2+x+1) + x+5
        = q(x)(x+1)^2 + x-1
        = [q(x)(x+1)+1](x+1)-2
設
   f(x) = r(x)(x^2+x+1)(x+1) + ax^2+bx+c
則
   f(x) = [r(x)(x+1)+a](x^2+x+1) + (b-a)x+(c-a)
∴ b-a = 1, c-a = 5
∴ f(x) = r(x)(x^2+x+1)(x+1) +ax^2+(a+1)x+(a+5)
        = [r(x)(x^2+x+1)+ax+1](x+1) + (a+4)
∴ a+4 = -2, a = -6, b = -5, c = -1
所以, f(x) 除以 (x^2+x+1)(x+1) 餘式為 -6x^2-5x-1

f(x)=q(x)(x²+x+1)(x+1)²+a(x²+x+1)(x+1)+b(x²+x+1)+(x+5)
f(-1)=-1-1
b(1-1+1)+(-1+5)=-2
b=-6
f(x)除以(x^2+x+1)(x+1)餘式為-6(x²+x+1)+(x+5)=-6x²-5x-1