Let P(x)=x4+ax3+bx2+cx+dbe a polynomial such that P(1)=1,P(2)=8,P(3)=27,P(4)=64,then the value of P(5) is_________.

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answer is 149.

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Detailed Solution

Clearly, P(x)−x3=0 has roots 1,2,3,4∴ P(x)−x3=(x−1)(x−2)(x−3)(x−4)⇒ P(x)=(x−1)(x−2)(x−3)(x−4)+x3Hence, P(5)=1×2×3×4+125=149.

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