Remainder Theorem

Let Ax + B be the remainder, then

x12 - 6x7 + 5 = Q(x) (x - 1)2 + Ax + B where Q(x) is some polynomial in x.

Sub x = 1:

1 - 6 + 5 = A + B

A + B = 0 ... (1)

Taking differentiation w.r.t. x:

12x11 - 42x6 = 2Q(x) (x - 1) + Q(x) (x - 1)2 + A

Sub x = 1:

12 - 42 = A

A = -30

B = 30

Hence the remainder is - 30x + 30

2012-11-09 15:12:32 補充：
After double checking, it is found that the method should be like this:

Since x^12 - 6x^7 + 5 is divisible by x - 1, we should perform division so that:

x^12 - 6x^7 + 5 = (x - 1)(x^11 + x^10 + x^9 + x^8 + x^7 - 5x^6 - 5x^5 - 5x^4 - 5x^3 - 5x^2 - 5x - 5)

2012-11-09 15:13:06 補充：
Hence the remainder should be the same as when x^11 + x^10 + x^9 + x^8 + x^7 - 5x^6 - 5x^5 - 5x^4 - 5x^3 - 5x^2 - 5x - 5 is divided by (x - 1), which is -30 by directly substituting x = 1 into x^11 + x^10 + x^9 + x^8 + x^7 - 5x^6 - 5x^5 - 5x^4 - 5x^3 - 5x^2 - 5x - 5

B=-A
x^12 -6x^7+5 = (x -1)^2 Q(x) +Ax -A
x^12 -6x^7+5 = (x -1)^2 Q(x) +A(x -1)
Divide both sides by (x-1),
x^11 +x^10 +x^9 +x^8 +x^7 -5x^6 -5x^5 -5x^4 -5x^3 -5x^2 -5x -5 = (x -1) Q(x) +A
Put x=1,
30 = (1 -1) Q(1) +A
A = -30
B = 30
The remainder is -30x +30

x^12 - 6x^7 + 5 = (x^2 - 2x + 1)Q(x) + R(x)

As x^2 = 2x - 1, so,
x^3 = 2x^2 - x = 3x - 2,
x^4 = 3x^2 - 2x = 4x - 3
x^5 = 4x^2 - 3x = 5x - 4
. . .
x^7 = 7x - 6
. . .
x^12 = 12x - 11
Therefore, the remainder R(x) is :
x^12 - 6x^7 + 5
= 12x - 11 - 6(7x - 6) + 5
= 12x - 11 - 42x + 36 + 5
= -30x + 30

1.
現在兩個答案都用 differentiation，若果未學過 differentiation，就不可以解這 Remainder theorem 的題目嗎？

2.
(-30x + 30) ÷ (x - 1)²
= -30(x - 1) ÷ (x - 1)²
= -30 ÷ (x - 1)

那麼，答案是 -30x + 30 還是 -30？

圖片參考：http://imgcld.yimg.com/8/n/HA00103546/o/20121109124507.jpg


睇下幫唔幫到你啦~~

2012-11-10 01:15:57 補充：
樓主：

可以照做長除法Long Division

不能這樣，餘式=-30x+30，不可抽完再約，因為分數不計的~~

2012-11-10 01:19:03 補充：
例：

20 / 6 = 3 ...2

不可變作
20 / 6 = 10 / 3 = 3 ...1

是不可以這樣做的

2012-11-12 10:11:02 補充：
Method 2:
Same as method 1 until A+B=0
B=-A
x^12 -6x^7+5 = (x -1)^2 Q(x) +Ax -A
x^12 -6x^7+5 = (x -1)^2 Q(x) +A(x -1)
Divide both sides by (x-1),
x^11 +x^10 +x^9 +x^8 +x^7 -5x^6 -5x^5 -5x^4 -5x^3 -5x^2 -5x -5 = (x -1) Q(x) +A
Put x=1,
30 = (1 -1) Q(1) +A
A = -30
B = 30
The remainder is -30x +30