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Let f(x) = x³ + px² +qx+2, g(x) = x³ + qx²+px-6
By factor theorem,
f(2) = 8 + 4p + 2q + 2 = 0
4p + 2q = -10 ... equation 1
g(2) = 8 + 4q + 2p - 6 = 0
2p + 4q = -2
p + 2q = -1 ... equation 2
equation 1 - 2 : 3p = -9
therefore, p = -3
Sub p = -3 into equation 2
-3 + 2q = -1
2q = 2
q = 1

Ans. p = -3, q = 1

Let f(x) = x³ + px² +qx+2, g(x) = x³ + qx²+px-6
By factor theorem,
f(2) = 8 + 4p + 2q + 2 = 0
4p + 2q = -10 ... (1)
g(2) = 8 + 4q + 2p - 6 = 0
2p + 4q = -2
p + 2q = -1 ... (2)
(1)-(2): 3p = -9
therefore, p = -3
Sub p = -3 into (2)
-3 + 2q = -1
2q = 2
q = 1

Ans. p = -3, q = 1

設F(x)為x³+px² +qx+2，G(x)為x³+qx²+px-6 。
因為x-2 是x³+px² +qx+2 and x³+qx²+px-6 的公因式，
F(2)=0 與 G(2)=0。
2³+p(2²)+q(2)+2=0
4p+2q=-10 (i)
2³+q(2²)+p(2)-6=0
4q+2p=-2
p+2q=-1(ii)

(i)-(ii) 3p=-9
　　　p=3

將p=3代入(ii)
p+2q=-1
3+2q=-1
2q=-4
q=-2

所以p=2，q=3。

2006-11-06 18:38:09 補充：
所以p=-2，q=-3。

2006-11-06 18:39:26 補充：
這個才是正確答案﹕(i)-(ii) 3p=-9　　　p=-3將p=-3代入(ii)p+2q=-1-3+2q=-12q=2q=1所以p=-3，q=1。