中四野 餘式定理 有3題好難

1.P(x) = (x - 1)(x+2)Q(x) + Ax+B
根據餘式定理,
P(1) = -4
A+B = -4 ...(1)
P(-2) = -28
-2A + B = -28 ...(2)
(1) - (2)：
3A = 24
A = 8 ....(3)
將(3)代入(1)，
8+B = -4
B = -12
A = 8, B = -12
∴餘式是8x - 12

2.(a)f(x)除以(x+1)時，餘數是-1
f(-1) = 1
(-1)99 +k = 1
-1 + k = 1
k = 2

(b)999+2 = 10Q+1
999= 10Q - 1
999= 10Q+10 - 10 - 1
999=10(Q - 1) + 9
∴所得的餘數是9

3.(x3+2ax﹣b)/(x﹣3)=m...1 <--設m係個整數的答案
x3+2ax﹣b=(x﹣3)m+1 <--佢未除個數o個陣 應該係個ans乘被除數 +番個餘數
x3+2ax﹣1=(x﹣3)m+b
(x3+2ax﹣1)/(x﹣3) = m...b
多項式是x3+2ax﹣1


2007-10-25 17:45:23 補充：
(ax＋b)是假設出來的餘式將結果A = 8及B = -12代入(ax＋b)中變成(8x - 12)

2007-10-25 21:03:18 補充：
f(x) = p(x)‧Q(x)＋R(x)其中R(x)是餘式所以9^99 =10(Q - 1)＋9 中，所得的餘數是9

a) f(1) = -4 and f(-2) = -28
Let the remainder of P(x) divided by (x-1)(x+2) be ax+b.
Then, f(1) = -4 = a + b ------------------(1)
f(-2) = -28 = -2a + b -------------(2)
(1) - (2), we have 24 = 3a, or a = 8.
By substituting a into (1), we have b = -4 - 8 = -12
Therefore the required remainder expression is 8x-12.

2a) f(-1) = (-1)^99 + k = 1
=&gt; k = 2

2b) From 1, when x = 9, it means that the remainder of (9^99 + 2) divided by (9+1) = 10 is 1. Therefore, the only possible last digit for 9^99 is 9 so that the last digit of 9^99 + 2 is 1.
Therefore, the remainder of 9^99 divided by 10 is 9.

3. 已知當x^3+2ax-b除以x-3時,所得的餘數是1 ,由此,試寫出一個以x為變數的三次多項式,
使它除以x-3時,所得的餘數為b .
Let g(x) = f(x) + c such that g(3) = b
=&gt; g(3) = f(3) + c
=&gt; b = 1 + c or c = b-1
g(x) = x^3+2ax+b+(b-1)
= x^3+2ax+2b-1