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📅 May 11, 2026 👤 scraface

May 11, 2026

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So $ g(x) $ has roots at $ x = 1, 2, 3, 4 $. But $ g(x) $ is a cubic p...
Therefore, such a cubic polynomial cannot exist — unless the values fo...
Then $ g(1) = g(2) = g(3) = g(4) = 0 $. So $ g(x) $ has roots at $ 1,2...
Unless — the only way this is possible is if $ g(x) $ is not degree 4...
So the only resolution is that our assumption of cubicity is violated...
Ah — perhaps we must use finite differences.
But the problem says $ f(x) $ is cubic. So either the problem is flawe...
Unless $ f(x) - 10x $ has four zeros — impossible for degree 3. So the...
Wait — perhaps the function is cubic but satisfies these values — but...
But here: $ f(1)=10, f(2)=20, f(3)=30, f(4)=40 $ — these lie on $ f(x)...

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