math 中五---polynomials

Let the remainder when p(x) is divided by (x+1)(x+2) be ax + b :thenP(x) = Q(x) * (x+1) (x+2) + ax + bBy remainder theoream :P(- 1) = - 7
0 + a(-1) + b = - 7
a - b - 7 = 0 .....(1)P(- 2) = - 25
0 + a(-2) + b = - 25
2a - b - 25 = 0 .....(2)(2) - (1) :a - 25 + 7 = 0
a = 18
Sub into (1) :
18 - b - 7 = 0
b = 11Answer : 18x + 11
2011-03-20 12:31:30 補充
Alternative :

P(x) = Q(x) * (x+1)(x+2) + k(x+1) - 7

P(-2) = 0 + k(-2+1) - 7

- 25 = - k - 7

k = 18

The remainder is k(x+1) - 7 = 18(x+1) - 7 = 18x + 11

2011-04-26 18:53:58 補充：
When a polynomial P(x) is divided by x+1, the remainders are -7:
=> P(-1)=-7...(1)

When a polynomial P(x) is divided by x+2, the remainders are -25:
=> P(-2)=-25...(2)

2011-04-26 18:54:05 補充：
When P(x) is divided by (x+1)(x+2), let the remainder be (ax+b):
P(x)=(x+1)(x+2)Q(x)+(ax+b)...(*)

Sub (1) into (*): P(-1)=-a+b=-7...(3)
Sub (2) into (*): P(-2)=-2a+b=-25...(4)

(3)-(4): a=18 and b=11

Hence, the remainder is 18x+11

2011-04-26 20:33:56 補充：
由題意 : P(x) = H(x) * (x+1) - 7 ,

而 P(x) = Q(x) * (x+1)(x+2) + k(x+1) - 7

可書為

P(x) = Q(x) * (x+2+k) (x+1) - 7

那麼 H(x) = Q(x) * (x+2+k)

;

當然也可設 P(x) = Q(x) * (x+1)(x+2) + k(x+2) - 25 來解。