maths!

f(x)=x3+px+q

By remainder theorem,
f(-1) = -3
(-1)^3 + p(-1) + q = -3
-1 - p + q = -3
-p + q = -2 ------------- (1)

Also,
f(-2) = 0
(-2)^3 + p(-2) + q = 0
-8 - 2p + q = 0
-2p + q = 8 ------------- (2)

(2) - (1):
(-2p + q) - (-p + q) = 8-(-2)
-p = 10
p = -10

sub p = -10 into (1)
-(-10) + q = -2
10 + q = -2
q = -12

2008-04-24 16:08:35 補充：
OK, 再做多次
f(x)=x^2+px+q

By remainder theorem,
f(-1) = -3
(-1)^2 + p(-1) + q = -3
1 - p + q = -3
-p + q = -4 ------------(1)

Also,
f(-2) = 0
(-2)^2 + p(-2) + q = 0
4 - 2p + q = 0
-2p + q = -4 -----------(2)

2008-04-24 16:08:48 補充：
(2) - (1):
(-2p + q) - (-p + q) = -4-(-4)
-2p + q + p - q = -4+4
-p = 0
p = 0

sub p = 0 into (1)
-(0) + q = -4
q = -4

Finished.

---------------
check:
f(x)=x^2 - 4 = (x+1)(x-1) - 3
f(x) = x^2 - 4 = (x+2)(x-2) + 0

so the answer should be correct

p=-6,q=-6
f(-1)=3(-1)-p+q=-3
f(-2)=3(-2)-2p+q=0


方法如上...之前打左steps..唔小心del左..
有問題搵我.
我估無計錯,combine埋就可以