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/ p(3) &= 27a + 9b + 3c + d = 21, \\
p(3) &= 27a + 9b + 3c + d = 21, \\
February 22, 2026
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Solution: Let $ p(x) = ax^3 + bx^2 + cx + d $. We are given:
p(1) &= a + b + c + d = 3, \\
p(2) &= 8a + 4b + 2c + d = 10, \\
p(4) &= 64a + 16b + 4c + d = 36.
We solve this system step by step. Subtract consecutive equations:
Now subtract Eq 1 from Eq 2:
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