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Home » So, \( r = \frac12\sqrt36 = 2\sqrt3 \). - jntua results

So, \( r = \frac12\sqrt36 = 2\sqrt3 \). - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 Now use $ p(1) $: $ 0 + 2 + 1 + d = 3 \Rightarrow d = 0 $.

📌 Thus, $ p(x) = 2x^2 + x $. Then $ p(0) = 0 $.

📌 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.

📌 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 + x + 0 $, and $ p(0) = 0 $.

📌 But let's verify all values:

📌 $ p(1) = 2 + 1 = 3 $,

📌 $ p(2) = 8 + 2 = 10 $,

📌 $ p(3) = 18 + 3 = 21 $,

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