A math ( formula)
回答 (3)
2006-12-08 12:31 am
✔ 最佳答案
p^3-p^2-p+1=p^2(p-1)-(p-1)=(p^2-1)(p-1)
=(p+1)(p-1)(p-1)
=(p+1)(p-1)^2
2006-12-07 16:33:21 補充：
樓主既題目已經俾錯左answer，並非(p-1^2)(p+1)，係(p-1)^2(p+1)。
2006-12-07 9:46 pm
Set f(p) = p^3 - p^2 - p + 1
We found that f(1) = 0 and f(-1) = 0.
By remainder theorem, (p+1) and (p-1) are factor of f(p).
Divide f(p) by (p+1),
we have f(p) / (p+1) = p^2 - 2p + 1 ==> f(p) = (p+1)(p^2 - 2p + 1)
Divide (p^2 - 2p + 1) by (p-1),
we have (p^2 - 2p + 1) / (p-1) = (p-1) ==> (p^2 - 2p + 1) = (p-1)^2
Therefore, f(p) = (p+1)(p-1)^2
We found that f(1) = 0 and f(-1) = 0.
By remainder theorem, (p+1) and (p-1) are factor of f(p).
Divide f(p) by (p+1),
we have f(p) / (p+1) = p^2 - 2p + 1 ==> f(p) = (p+1)(p^2 - 2p + 1)
Divide (p^2 - 2p + 1) by (p-1),
we have (p^2 - 2p + 1) / (p-1) = (p-1) ==> (p^2 - 2p + 1) = (p-1)^2
Therefore, f(p) = (p+1)(p-1)^2
2006-12-07 9:42 pm
p^3 - p^2 - p + 1
=p^2(p-1)-(p-1)
=(p^2-1)(p-1)
=[(p-1)(p+1)](p-1)
=(p-1)^2(p+1)
=p^2(p-1)-(p-1)
=(p^2-1)(p-1)
=[(p-1)(p+1)](p-1)
=(p-1)^2(p+1)
收錄日期: 2021-04-12 18:30:19
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