餘式定理高次方 x^88+x^27-2x^10-3x^5+x 除以 x^3+2x^2+2x+1的餘式為? 沒有答案,麻煩各位高手了!!?

2015-11-01 3:22 pm

回答 (3)

2015-11-01 10:22 pm
✔ 最佳答案
Sol
x^3+2x^+2x+1=(x+1)(x^2+x+1)
設f(x)=x^88+x^27-2x^10-3x^5+x=q(x)(x+1)(x^2+x+1)+a(x^2+x+1)+bx+c
w為x^2+x+1=0之一根
(w-1)(w^2+w+1)=0
w^3-1=0
w^3=1
f(w)=w^88+w^27-2w^10-3w^2+w=bw+c
w+1-2w-3w^2+w=bw+c
3w^2+bw+(c-1)=0
b=3,c=4
f(x)=x^88+x^27-2x^10-3x^5+x=q(x)(x+1)(x^2+x+1)+a(x^2+x+1)+3x+4
f(-1)=1-1-2+3-1=a(1-1+1)-3+4
0=a+1
a=-1
餘式=-(x^2+x+1)+3x+4=-x^2+2x+3
2015-11-02 1:39 am
因為 x³ + 2x² + 2x + 1 = (x + 1)(x² + x + 1)
假設 x⁸⁸ + x²⁷ - 2x¹⁰ - 3x⁵ + x 除以 (x + 1)(x² + x + 1) 的商是 q(x),餘式是 ax² + bx + c,則
x⁸⁸ + x²⁷ - 2x¹⁰ - 3x⁵ + x = (x + 1)(x² + x + 1)q(x) + ax² + bx + c
==> (x⁸⁸ + x²⁷ - 2x¹⁰ - 3x⁵ + x)(x - 1) = (x + 1)(x - 1)(x² + x + 1)q(x) + (ax² + bx + c)(x - 1) ⋯⋯ ①
==> x⁸⁹ - x⁸⁸ + x²⁸ - x²⁷ - 2x¹¹ + 2x¹⁰ - 3x⁶ + 3x⁵ + x² - x = (x + 1)(x³ - 1)q(x) + ax³ - ax² + bx² - bx + cx - c ⋯⋯ ②
代 x = -1 入 ① 式,得
1 - 1 - 2 + 3 - 1 = 0 + a - b + c
==> a - b + c = 0 ⋯⋯ ③
代 x³ = 1 入 ② 式,得
x² - x + x - 1 - 2x² + 2x - 3 + 3x² + x² - x = 0 + a - ax² + bx² - bx + cx - c
==> 3x² + x - 4 = (b - a)x² + (c - b)x + (a - c)
所以 b - a = 3、c - b = 1,代入 ③ 式,得 a = -1
即 a= -1, b = 2, c = 3
答案:餘式是 (-x² + 2x + 3)
2015-11-01 4:04 pm
參考可能錯
x^3+2x^2+2x+1=(x^3+1)+2x(x+1)=(x+1)(x^2+x+1)

x^3-1=(x-1)(x^2+x+1)=0 ==>x^3=1

x^88+x^27-2x^10-3x^5+x=(x^3)^29*x+(x^3)^9-2(x^3)^3*x-3(x^3)*x^2+x
=x+1-2x-3x^2+x=-3x^2+1=-3(-1)^2+1= -2


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