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Home » "This Bam BemoL-Down Chaos Will Transform How You Play Guitar Instantly! - Abbey Badges

"This Bam BemoL-Down Chaos Will Transform How You Play Guitar Instantly! - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Substitute $ a = 0 $ into (A): $ 6(0) + b = 2 \Rightarrow b = 2 $.

📌 Substitute $ a = 0, b = 2 $ into Eq 1:

📌 7(0) + 3(2) + c = 7 \Rightarrow 6 + c = 7 \Rightarrow c = 1.

📌 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:

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