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Webseite an der Universität Wrocław - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Solution: Since $ p(x) $ is a cubic polynomial, when dividing by $ x^4 - 1 $, the remainder $ r(x) $ must be a polynomial of degree less than 4. However, because $ x^4 - 1 $ has roots $ \pm 1, \pm i $, and $ p(x) $ is real, the remainder can be expressed as any cubic or lower, but here we are to find the **actual remainder**, which is unique and of degree $ < 4 $. But since $ p(x) $ is already degree 3, and division by degree 4 polynomial yields a unique remainder of degree $ < 4 $, and since $ p(x) $ agrees with any degree ≤3 polynomial at these 4 points, we conclude:

📌 p(x) = \text{remainder of } p(x) \div (x^4 - 1) \quad \text{only if } \deg p < \deg(x^4 - 1)

📌 But $ p(x) $ is cubic, so remainder $ r(x) = p(x) $, but only if $ x^4 - 1 $ divides $ p(x) - r(x) $, which is not true unless $ p(x) $ is degree ≤3 — which it is.

📌 Wait: actually, **any** polynomial $ p(x) $ of degree less than 4 **is** the remainder upon division by $ x^4 - 1 $. Since $ \deg(p) = 3 < 4 $, the division algorithm gives:

📌 p(x) = 0 \cdot (x^4 - 1) + p(x)

📌 So the remainder is **$ p(x) $** itself. But the problem asks for the remainder — which is a polynomial of degree less than 4 — so it is $ p(x) $, but we are to **report it in standard form**.

📌 But we are given four values, so we reconstruct $ p(x) $.

📌 Let $ p(x) = ax^3 + bx^2 + cx + d $

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