maths 績多項式

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

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