Question

求f(x)=x^1999-1除以x^4+x^3+2x^2

Answer 1

x^4+x^3+2x^2+x+1==(x^2+x+1)(x^2+1)

x³+2x²+2x+1=(x³+1)+2x(x+1)=(x+1)(x²+x+1)

1) f(x)除以x+1的余式为f(-1)=0

2) (x-1)(x²+x+1)=x³-1

f(x)=x*(x³)^9+(x³)^9-2*x(x³)³-3x³*x²+x除以x³-1的余式为

x+1-2x-3x²+x=-3x²+1 (以x³=1代入f(x))

所以f(x)=(x-1)(x²+x+1)g(x)-3x²+1

````````=(x²+x+1)[(x-1)g(x)]-3(x²+x+1)+3x+4

被x²+x+1除的余式为3x+4

3) 令f(x)=(x³+2x²+2x+1)Q(x)+ax²+bx+c

`````````=(x+1)(x²+x+1)Q(x)+a(x²+x+1)+3x+4

由1),f(-1)=0 => f(-1)=a-3+4=0 => a=-1

故所求余式为-(x²+x+1)+3x+4=-x²+2x+3