求f(x)=x^1999-1除以x^4+x^3+2x^2+x+1的余式?f(x)=x^88+x^27-2x^10-3x^5+x除以x^3+2x^2+2x+1的余式是?

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求f(x)=x^1999-1除以x^4+x^3+2x^2+x+1的余式?f(x)=x^88+x^27-2x^10-3x^5+x除以x^3+2x^2+2x+1的余式是?
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求f(x)=x^1999-1除以x^4+x^3+2x^2+x+1的余式?f(x)=x^88+x^27-2x^10-3x^5+x除以x^3+2x^2+2x+1的余式是?
求f(x)=x^1999-1除以x^4+x^3+2x^2
+x+1的余式?
f(x)=x^88+x^27-2x^10-3x^5+x除以x^3+2x^2+2x+1的余式是?

求f(x)=x^1999-1除以x^4+x^3+2x^2+x+1的余式?f(x)=x^88+x^27-2x^10-3x^5+x除以x^3+2x^2+2x+1的余式是?
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

令f(x)=q(x)g(x)+r(x),求出g(x)=0的根a,带入前式,得f(a)=r(a).根据拉格朗日插值公式即可求出r(x).

参考:

http://tieba.baidu.com/p/919306508

http://www.docin.com/p-488547402.html

http://www.cqvip.com/Read/Read.aspx?id=33323480