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Home » This absolute value equation splits into two linear equations: - All Square Golf

This absolute value equation splits into two linear equations: - All Square Golf

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 f(x+y) + f(x-y) = c(x+y)^2 + c(x-y)^2 = c(x^2 + 2xy + y^2 + x^2 - 2xy + y^2) = c(2x^2 + 2y^2) = 2cx^2 + 2cy^2 = 2f(x) + 2f(y)

📌 So it satisfies the equation.

📌 Given $ f(1) = c(1)^2 = 3 \Rightarrow c = 3 $, so $ f(x) = 3x^2 $.

📌 Then $ f(5) = 3 \cdot 25 = \boxed{75} $

📌 Question: A cartographer uses a polynomial $ p(x) $ of degree 3 such that $ p(0) = 2 $, $ p(1) = 5 $, $ p(2) = 12 $, and $ p(3) = 31 $. Find the remainder when $ p(x) $ is divided by $ x^4 - 1 $.

📌 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.

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