F.4 Maths...急急急!!!

您好，我是 lop，高興能解答您的問題。

(1.1)

A(x+2) + B(x-3) ≡ x + 12
Ax + Bx + 2A - 3B ≡ x + 12
(A+B)x + (2A-3B) ≡ x + 12

A + B = 1 --- (1)
2A - 3B = 12 --- (2)

From (1) :

A = 1 - B --- (3)

Sub. (3) into (2)

2(1-B) - 3B = 12
2 - 2B - 3B = 12
-5B = 10
B = -2 --- (4)

Sub. (4) into (3) :

A = 1 - (-1)
A = 2

Answer : A = 2 , B = -2

(1.2)

x^3 - x^2 - 6x - 4 + 2x + 3 ≡ x^3 + ax^2 + bx + c
x^3 - x^2 - 4x - 1 ≡ x^3 + ax^2 + bx + c

Answer : a = -1 , b = -4 , c = -1

(2.1)

f(x) = 2x^3 - 3x^2 + kx + 6
= (2x^3+1x^2) - 4x^2 + kx + 6
= (2x^2)(2x+1) - (4x^2+2x) + (k+2)x + 6
= (2x^2)(2x+1) - 2x(2x+1) + (k+2)x + 6

k+2 = 2p
6 = p
k = 10

Answer : k = 10 .

2011-12-19 11:26:12 補充：
corr. of 001 ,

Sub. (4) into (3) :

A = 1 - (-2)
A = 3

Answer : A = 3 , B = -2

lop在第2題並不是使用餘式定理

1.
A(x+2)+B(x-3)≡x+12
在=AX+2A+BX-3B
在=(A+B)x+(2A-3B)
A+B=1 ...i
2A-3B=12 ..ii
2*i,
2A+2B=2 ...iii
ii-iii
-5B=10
B=-2
代B=-2入i
A-2=1
A=1
所以
(A,B)=(1,-2)

2.
x^3+ax^2+bx+c =(x^2-2x-4)(x+1)+2x+3
右=x^3-2x^2+4x+x^2-2x+4+2x+3
右=x^3- x^2+4x+7
所以
a=-1
b=4
c=7

1.
f(x)=2x^3-3x^2+kx+6
f(-1/2)=2(-1/2) ^3-3(-1/2) ^2+k(-1/2) +6
0=5-k/2
-5=-k/2
k=10

1.
Ax + By + C ≡ Px + Qy + R
=> A = P, B = Q, C = R

2.
when 7 is divided by 2, what do you get?
7 = 2 (3) + 1
when f(x) is divided by g(x)
f(x) = g(x) Q(x) + R(x) where the degree of R(x) < degree of g(x)
x^3+ax^2+bx+c = (x^2-2x-4) (x+1) + (2x+3)