Let P ( x ) = x 5 + a x 4 + b x 3 + c x 2 + d x + e be a polynomial with real coefficients and satisfying the property P ( n ) = 1 0 n for n = 1 , 2 , 3 , 4 , and 5 . Find the value of a + b + c + d + e .
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Let g ( x ) = P ( x ) − 1 0 x .Then g ( x ) = 0 for x = 1 , 2 , 3 , 4 , 5 .As P ( x ) is monic,therefore g ( x ) is monic as well.
Therefore, g ( x ) can be represented as: ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) = P ( x ) − 1 0 x ⟹ P ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) + 1 0 x ⟹ P ( x ) = x 5 − 1 5 x 4 + 8 5 x 3 − 2 2 5 x 2 + 2 8 4 x − 1 2 0 ⟹ a = − 1 5 , b = 8 5 , c = − 2 2 5 , d = 2 8 4 , e = − 1 2 0 ⟹ a + b + c + d + e = 9
Instead of expanding the polynomial, we can substitute in x = − 1 .
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