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📅 May 4, 2026 👤 scraface

May 04, 2026

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🔥 Popular Posts

p(x) = 0 \cdot (x^4 - 1) + p(x)
So the remainder is **$ p(x) $** itself. But the problem asks for the...
But we are given four values, so we reconstruct $ p(x) $.
Let $ p(x) = ax^3 + bx^2 + cx + d $
Use the values:
$ p(0) = d = 2 $
$ p(1) = a + b + c + d = 5 \Rightarrow a + b + c = 3 $
$ p(2) = 8a + 4b + 2c + d = 12 \Rightarrow 8a + 4b + 2c = 10 $
$ p(3) = 27a + 9b + 3c + d = 31 \Rightarrow 27a + 9b + 3c = 29 $
From (2): $ a + b + c = 3 $

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