Solution: $ d(t) = pt^3 + qt^2 + rt + s $. Compute $ d'(t) = 3pt^2 + 2qt + r $. From $ d(1) = p + q + r + s = 10 $, $ d'(1) = 3p + 2q + r = 12 $, $ d(2) = 8p + 4q + 2r + s = 28 $, $ d'(2) = 12p + 4q + r = 30 $. Subtract first equation from third: $ 7p + 3q + r = 18 $. Subtract $ d'(1) $ from this: $ (7p + 3q + r) - (3p + 2q + r) = 4p + q = 6 $. From $ d'(2) $: $ 12p + 4q + r = 30 $, and $ d'(1) $: $ 3p + 2q + r = 12 $. Subtract: $ 9p + 2q = 18 $. Now solve $ 4p + q = 6 $ and $ 9p + 2q = 18 $. Multiply first by 2: $ 8p + 2q = 12 $. Subtract: $ p = 6 $. Then $ 4(6) + q = 6 $ â†’ $ q = -18 $. From $ d'(1) $: $ 3(6) + 2(-18) + r = 12 $ â†’ $ 18 - 36 + r = 12 $ â†’ $ r = 30 $. From $ d(1) $: $ 6 - 18 + 30 + s = 10 $ â†’ $ s = -8 $. Thus, $ d(0) = s = -8 $. Final answer: $ oxed{-8} $.

Sun, 03 May 2026 05:20:27 +0000

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