Polynomial 的问题

外星護法

When a polynomian f(x) is divided by (x-a) and (x-b), the remainder are f(a) and f(b) respectively.

Find the remainder when f(x) is divided by (x-a)(x-b).

请问要怎么做 ？？

Allmaths

回复 1# 外星護法


when f(x) divided by (x-a)(x-b), it gives a remainder Ax+B where A and B are constants.


f(x)=(x-a)(x-b)q(x)+Ax+B                  (where q(x) is the quotient)


when x=a,the remainder is

f(a)=Aa+B

when x=b,the remainder is

f(b)=Ab+B


差不多是酱的东西吧。。

不是很确定答案。。。

mathlim

When a polynomian f(x) is divided by (x-a) and (x-b), the remainder are f(a) and f(b) respectively.

Find the remainder when f(x) is divided by (x-a)(x-b).

请问要怎么做 ？？

方法一：
设 f(x) = (x-a)(x-b)q(x) + Ax+B
f(a) = Aa+B ——(1)
f(b) = Ab+B ——(2)
(1) - (2)
A(a-b) = f(a)-f(b)
A = [f(a)-f(b)]/(a-b)
B = [af(b)-bf(a)]/(a-b)
∴ 所求余式为 {[f(a)-f(b)]x + [af(b)-bf(a)]}/(a-b)。

方法二：
设 f(x) = (x-a)(x-b)q(x) + A(x-a)+f(a)
f(b) = A(b-a)+f(a)
A = [f(a)-f(b)]/(a-b)
∴ 所求余式为 [f(a)-f(b)](x-a)/(a-b) + f(a)
即 {[f(a)-f(b)]x + [af(b)-bf(a)]}/(a-b)。

方法三：
设 f(x) = (x-a)(x-b)q(x) + A(x-b)+f(b)
f(a) = A(a-b)+f(b)
A = [f(a)-f(b)]/(a-b)
∴ 所求余式为 [f(a)-f(b)](x-b)/(a-b) + f(b)
即 {[f(a)-f(b)]x + [af(b)-bf(a)]}/(a-b)。

外星護法

原来是这样~ 明白了~ 谢谢指教~~