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📰 So $ g(x) $ has roots at $ x = 1, 2, 3, 4 $. But $ g(x) $ is a cubic polynomial, while a cubic polynomial cannot have 4 distinct roots unless it is identically zero — contradiction.
📰 Therefore, such a cubic polynomial cannot exist — unless the values force a special form. But the values $ f(1)=10, f(2)=20, f(3)=30, f(4)=40 $ suggest $ f(x) - 10x $ has four zeros. So define:
📰 Then $ g(1) = g(2) = g(3) = g(4) = 0 $. So $ g(x) $ has roots at $ 1,2,3,4 $, so $ g(x) = k(x-1)(x-2)(x-3)(x-4) $, but this is degree 4 — contradiction, since $ f(x) $ is cubic.
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