Re: [中學] 多項式餘式求法

[yueayase (scrya)]

※ 引述《hotplushot (熱加熱)》之銘言：
: 多項式 x^38-2x^26+3x^11-x
: 除以
: (1)x^3-1
  令 x^3 - 1 = 0 => x^3 = 1

     x^38-2x^26+3x^11-x

  ≡ x^2-2x^2+3x^2-x

  ≡ 2x^2-x

: (2)x^2+x+1

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

   令x^3-1 = 0

   x^38-2x^26+3x^11-x

  ≡ 2x^2-x

  再令 x^2+x+1 = 0 => x^2 = -x-1

  x^38-2x^26+3x^11-x

≡ 2(-x-1)-x

≡ -3x-2

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

     (x^4+x^2+1)(x^2-1) = x^6 - 1

   x^38-2x^26+3x^11-x

≡ x^2-2x^2+3x^5-x

≡ 3x(-x^2-1)-x^2-x

≡ -3x^3-x^2-4x



: (4)x^5+x^3+x^2+1

     x^5+x^3+x^2+1 = x^3(x^2+1)+(x^2+1) = (x^3+1)(x^2+1)

     令 x^3 = -1
     => x^38-2x^26+3x^11-x ≡ x^2-2x^2-3x^2-x ≡ -4x^2-x

     令 x^2 = -1
     => x^38-2x^26+3x^11-x ≡ -1-2(-1)+3(-x)-x ≡ -4x+1

     則 x^38-2x^26+3x^11-x = (x^3+1)Q1(x) + (-4x^2-x)

                           = (x^2+1)Q2(x) + (-4x+1)


     令 x^38-2x^26+3x^11-x = (x^3+1)(x^2+1)Q3(x) + (x^3+1)(ax+b) + (-4x^2-x)

     (因為被5次式除餘式最高為4次)

     同除x^2+1

     => -4x+1 ≡ (x^3+1)(ax+b) + (-4x^2-x)

              ≡ (-x+1)(ax+b) + (4-x)

              ≡ -ax^2+(a-b-1)x+ (b+4)

              ≡ (a-b-1)x + (a+b+4)


     => a-b-1=-4, a+b+4 = 1

     => a = 0, b = -3

     故餘式 = (x^3+1)(ax+b) + (-4x^2-x) = -3(x^3+1) + (-4x^2-x)
                                        = -3x^3-4x^2-x-3


: 的餘式各為何??

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推 hotplushot  :感謝大大
推 TRAP        :最後一個 怪怪的......
→ yueayase    :的確是錯的
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※ 編輯: yueayase        來自: 111.251.160.15       (12/14 00:46)
推 JohnMash    :a=-3,b=0