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非零多項式f(x)可以寫成：

f(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_2 x² + a_1 x + a_0

各項係數和其實即是 f(1)。

留意 f(1) = a_n + a_(n-1) + ... + a_2 + a_1 + a_0

即現在 f(1) = 5。

題目求f(x)+3除以x−1的餘數。

設 g(x) = f(x) + 3，也為多項式。

g(x)除以x−1的餘數是 g(1)

g(1) = f(1) + 3 = 5 + 3 = 8

答案為 (D)。

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