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Boulder KE = 0 J - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solution: We are given a cubic polynomial \( h(x) = x^3 + px + q \), a...
h(1) = 1^3 + p(1) + q = 1 + p + q = 4 \quad ext{(1)}
h(2) = 8 + 2p + q = 10 \quad ext{(2)}
h(3) = 27 + 3p + q = 24 \quad ext{(3)}
p + q = 3 \quad ext{(i)}
From (2):
2p + q = 2 \quad ext{(ii)}
+ q = 3 \Rightarrow q = 4
Now verify with (3):
h(3) = 27 + 3(-1) + 4 = 27 - 3 + 4 = 28

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