Re: [中學] 餘式定理
f(x) = (x - 1)^3 [px + b] + 3
= (x - 2 + 1)^3 [px + b] + 3
= (x - 2)*... + [px + b + 3]
= (x - 2)*... + b + 3 + 2p
=> b + 2p = 3
同理
f(x) = (x + 2 - 3)^3 [px + b] + 3
= (x + 2)*... [-27px + (3 - 27b)]
= (x + 2)*... + [-27b + 54p + 3]
=> 2p - b = 1
=> b = 1
p = 1
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06/21 21:11, , 1F
06/21 21:11, 1F
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