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 +

["Understanding Degree ≤ 3 Polynomial Problems: Why Minimal Degree Often is Quadratic", "In many math olympiad and competitive problem sets, you’ll encounter questions asking for the interpolating polynomial of degree at most 3 given a few data points. A recurring insight—especially in problems phrased "quadratic or cubic"—is that the solution frequently reduces to a quadratic, even when higher-degree polynomials seem possible. But why is that?", "considen a set of points $(x_i, y_i)$ where the $y_i$ values fit a quadratic relationship. Contest problems often describe solutions as if cubic, yet no cubic terms are truly enforced. The key lies in degree bounded constraints: when asked for a polynomial of degree ≤ 3, the minimal-degree polynomial that interpolates the data might very likely be degree 2—because higher-degree terms aren’t forced or justified by the data.", "### The Quadratic Assumption: No Degree > 2 is Required", "If values collectively follow a quadratic trend—say fitting $ p(x) = ax^2 + bx + c $—then any cubic polynomial $ p(x) = ax^3 + bx^2 + cx + d $ satisfying those values must have $ a = 0 $. That’s because imposing a cubic term introduces unnecessary complexity when the data doesn’t demand it. The interpolating polynomial of minimal degree satisfying the conditions is thus quadratic, despite the problem’s cubic phrasing.", "### The Cubic Framework: Convenience Over Necessity", "Some problems define the solution space as polynomials of degree ≤ 3, leaving room for a zero cubic coefficient. For example, choosing:\n$$\np(x) = 0 \cdot x^3 + 2x^2 + x + 0\n$$\nthis cubic polynomial is identically $ 2x^2 + x $, and $ p(0) = 0 $. Here, although the degree is capped at 3, the leading coefficient vanishes—so the polynomial is effectively quadratic.", "This trick—to model the interpolant as degree ≤ 3 while effectively using lower-degree terms—is common. It acknowledges the formal constraint without forcing higher-degree behavior.", "### So, what does $ p(0) $ equal?", "Given that the smallest-degree polynomial satisfying the interpolation conditions is quadratic (due to no higher-degree forcing), and notational convenience allows writing it as a cubic, we write:\n$$\np(x) = 2x^2 + x \quad \Rightarrow \quad p(0) = 0\n$$\nEven when representing it as $ p(x) = 0x^3 + 2x^2 + x + 0 $, evaluating at $ x = 0 $ gives:\n$$\np(0) = 0 \cdot 0 + 2 \cdot 0 + 0 + 0 = 0\n$$", "### Final Takeaway", "When solving interpolation problems with degree ≤ 3:\n- Prioritize minimizing the degree—quadratic is often sufficient.\n- A cubic form with $ a = 0 $ is valid and simplifies evaluation.\n- Thus, $ p(0) $ is determined by the actual coefficients, typically yielding $ p(0) = 0 $ for data truly following quadratics.", "Next time a problem allows degree ≤ 3, recall: not all solutions require degree 3—often, the elegant and minimal solution is quadratic, and $ p(0) = 0 $ follows naturally."]