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 (degree at most 3), and we found $ a = 0 $, the polynomial is actually quadratic. But the problem says cubic, so we must reconsider. - go-checkin.com