Re: [中學] 多項式餘式求法

看板Math作者yueayase (scrya)時間14年前 (2011/12/13 01:19)推噓3(3推 0噓 2→)

留言5則, 4人參與討論串2/4 (看更多)

※ 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?? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.251.160.15

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hotplushot

12/13 01:24, , 1^F

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TRAP

12/13 11:43, , 2^F

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yueayase

12/14 00:32, , 3^F

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