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標題: 中4 maths come!!!!!!!!!! [打印本頁]

作者: 承b 時間: 23-3-2012 19:27
標題: 中4 maths come!!!!!!!!!!
when p(x)is divided by (x^2)+1, the remainder is 2x+3.
given p(0)=0
find the remainder when p(x) is divided by x((x^2)+1)

作者: 承b 時間: 23-3-2012 20:47

作者: 承b 時間: 23-3-2012 20:48

作者: 火中有火 時間: 23-3-2012 23:09
本帖最後由火中有火於 24-3-2012 22:22 編輯

Answer係咪 -3x^2+2x ?

你肯定條數係比左'x^2+1'你?
因為真係好怪
搞到我剩係識屈一個合理的p(x)出來試...

eg. p(x)=(x^2 + 1)(x^2 - 3) + (2x + 3)          ←符合p(x)=0
         =x^4 - 2x^2 + 2x
         =x(x^3 - 2x + 2)
         =(x^3 + x)x - 3x^2 + 2x

eg.p(x)=(x^2 + 1)(x - 3) + (2x + 3)             ←符合p(x)=0
         =x^3 - 3x^2 + 3x
         =(x^3 + x) - 3x^2 + 2x
..............................................................................................................

↓平時做開果d係咁的↓

The remainder is 9 when f(x) is divided by x-5 and
the remainder is -5 when f(x) is divided by x+2.
Find the remainder when f(x) is divided by (x+2)(x-5)

Let the remainder be px+q
f(5)=5p+q=9 and f(-2)=-2p+q=-5
...
p=2 and q=-1

The remainder is 2x+1

..............................................................................................................

如果係你打錯題目搞到我做咁耐的話我會殺左你

作者: 火中有火 時間: 24-3-2012 22:20
好了今日日討論完有正解

做 polynomial 的 long division 要知道一點
when the divider is of degree n, the highest possible degree of the remainder is n-1
ie. when the divider is of degree 3, the highest possible degree of the remainder is 2

我地又知道如果x-a係factor的話,咁p(a)=0
同埋the remainder is p(b) when p(x) is divided by (x-b)

呢條題目的divider係degree 3
我地可以 let the remainder be px^2+qx+r, where p,q and r are constants.

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p(0)=p*0^2+q*0+r=0
r=0

p(i)=pi^2+qi+r=2i+3
-p+qi+0=2i+3
-p+qi=3+2i
p=-3 and q=2

The remainder is -3x^2+2x when p(x) is divided by x^3+x.

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不過都係想講多次粗口...因為真係冇見過咁的題目...
粗口粗口粗口粗口粗口粗口粗口粗口粗口粗口.........

作者: 承b 時間: 25-3-2012 19:24

good

作者: Ron 時間: 22-5-2012 23:43
其實呢ｄ我好多唔得了

順手借位一下

作者: 23535350 時間: 16-6-2012 13:56

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