若
被除式是 x^4 + Ax^3 + Bx^2 - x + 15
除式是 x^2 + x - 2
商式是 x^2 - 4
餘式是 3x + 7
求 A 和 B 的值
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maths 績多項式
2006-11-07 1:26 am
回答 (2)
2006-11-07 2:21 am
✔ 最佳答案
若 被除式是 x^4 + Ax^3 + Bx^2 - x + 15
除式是 x^2 + x - 2
商式是 x^2 - 4
餘式是 3x + 7
求 A 和 B 的值,
x^4+Ax^3+Bx^2-x+15=(x^2+x-2)(x^2-4)+3x+7
x^4+Ax^3+Bx^2-x+15=x^4+x^3-2x^2-4x^2-4x+8+3x+7
x^4+Ax^3+Bx^2-x+15=x^4+x^3-6x^4-x+15
By comparing like terms,
A = 1
B = - 6
2006-11-06 18:22:27 補充：
= 等於恆等
參考: me
2006-11-07 1:34 am
x^4 + Ax^3 + Bx^2 - x + 15 = (x^2 + x - 2)(x^2 - 4)+3x+7
= x^4-4x^2+x^3-4x-2x^2+8+3x+7
= x^4+x^3-6x^2-x+15
So, A=1, B=-6
= x^4-4x^2+x^3-4x-2x^2+8+3x+7
= x^4+x^3-6x^2-x+15
So, A=1, B=-6
參考: myself
收錄日期: 2021-04-12 22:31:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061106000051KK02532