已知g(x)除以x-2的餘數為6,求g(x-2)除以x-4的餘數
為什麼是g(4-2)???
求g(x-2)除以x-4的餘數
2014-01-17 6:53 am
回答 (2)
2014-01-17 6:59 am
✔ 最佳答案
g(x) 除以 x-2 的餘數為 6 ，即是 g(2) = 6。求 g(x-2) 除以 x-4 的餘數。
為免混淆，你可以先令 h(x) = g(x-2)
那麼 求 g(x-2) 除以 x-4 的餘數，即是求 h(x) 除以 x-4 的餘數。
答案是 h(4)。
由於 h(x) = g(x-2)，
答案是 h(4) = g(4-2) = g(2) = 6。
2014-01-17 4:11 pm
g(x) = (x - 2) Q(x) + 6
Put x = y - 2
g(x) = g(y - 2) = [ (y - 2) - 2] Q(y - 2) + 6
g(y - 2) = ( y - 4) Q(y - 2) + 6
That is
g(x - 2) = (x - 4) Q(x - 2) + 6 since y is dummy variable.
By Remainder theorem, when divided by (x - 4), we put x = 4,
that is
g(4 - 2) = (4 - 4) Q(4 - 2) + 6 = 6
so the remainder is 6 and that explained by it is g(4 - 2).
Put x = y - 2
g(x) = g(y - 2) = [ (y - 2) - 2] Q(y - 2) + 6
g(y - 2) = ( y - 4) Q(y - 2) + 6
That is
g(x - 2) = (x - 4) Q(x - 2) + 6 since y is dummy variable.
By Remainder theorem, when divided by (x - 4), we put x = 4,
that is
g(4 - 2) = (4 - 4) Q(4 - 2) + 6 = 6
so the remainder is 6 and that explained by it is g(4 - 2).
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