多項式唔識做....thanks!

2007-03-29 4:49 am

1)
求當f(x)=x^2 - 2x 除以下各項時的餘數:
a.2x
b.x - 2a + 1

2)
求當f(x)=3x^2 + x -4 除以下各項時的餘數:
a.6x
b.x + 2 - 3a

3)
已知h(x) = x^3 - ax^2 - 3x + 2b。當h(x)除以x+1時,餘數為2;當f(x)除以x+2時,餘數為-8
a.求a和b的值
b.求當h(-2x)除以x+2時的餘數

回答 (1)

阿力仔

2007-03-29 7:19 am

✔ 最佳答案

1)
a. 餘數 = f(0)=(0)^2 - 2(0) = 0
b. 餘數 = f(2a - 1) = (2a-1)^2 - 2(2a-1) = (2a-1)(2a-1-2) = (2a-1)(2a-3)

2)
a. 餘數 = f(0) = 3(0)^2 + (0) - 4 = -4
b. 餘數 = f(3a-2) = 3(3a-2)^2 + (3a-2) - 4 = 3(9a^2-12a+4)+3a-6 = 27a^2-33a+6

3)
a. h(-1) = (-1)^3-a(-1)^2-3(-1)+2b = 2
-1-a+3+2b = 2
a-2b=0 .............(1)
h(-2) = (-2)^3-a(-2)^2-3(-2)+2b = -8
-8-4a+6+2b=-8
2a-b=3 .............(2)

2x(1)-(2)
2x(-2b)-(-b)=0-3
-3b = -3
b = 1
Put b into (1)
a = 2

b. h(x) = x^3 - 2x^2 - 3x + 2
h(-2*-2) = h(4) = (4)^3 - 2(4)^2 - 3(4) + 2 = 64-32-12+2 = 22

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