$ p(1) = 2 + 1 = 3 $,

February 22, 2026

$ p(1) = 2 + 1 = 3 $,

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But wait—this is quadratic, but the problem states $ p(x) $ is cubic. Contradiction? Not necessarily—$ a = 0 $ is allowed if the cubic coefficient is zero, but the problem says cubic, implying degree exactly 3. So we must assume it's cubic, but our solution gives $ a = 0 $. That means the data fits a quadratic, but we are told it's cubic. So either the model is misclassified, or we must accept the interpolating polynomial, regardless of degree. Since the interpolation yields a unique cubic (degr
Wait—perhaps the problem allows degree ≤ 3. Many contest problems phrase cubic loosely. Given the values fit a quadratic, and no higher-degree terms are forced, the minimal-degree interpolating polynomial is quadratic. Since the problem asks to find $ p(0) $, and the unique cubic polynomial (in degree ≤ 3) satisfying the values must have $ a = 0 $, we proceed with $ p(x) = 2x^2 + x $, so $ p(0) = 0 $. However, to ensure degree 3, suppose we include a zero cubic term. Then $ p(x) = 0x^3 + 2x^2 +
But let's verify all values:
$ p(2) = 8 + 2 = 10 $,
$ p(3) = 18 + 3 = 21 $,
$ p(4) = 32 + 4 = 36 $.

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