A Cubic Equation... or is it?

Algebra Level 4

Let P ( x ) P(x) be a monic quadratic polynomial satisfying P ( 1 ) = 1 , P ( 2 ) = 8 P(1)=1, P(2)=8 , and P ( 3 ) = 27 P(3)=27 . Find P ( 4 ) P(4) .

58 54 128 64

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Amogha Pokkulandra by Amogha Pokkulandra , 23, USA

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2 solutions

Notice that this P(x) is not cubic as stated by the problem. We can solve this as if it were a sequence. If we take the differences between 1 and 8, and 8 and 27, we get the numbers 7 and 19 because 8-1 and 27-8 is equal to 7 and 19 respectively. The difference between 7 and 19 is 12. If we add 12 again to 19, we get 31. If we add 31 to 27, we get 58 \boxed{58} .

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But doesn't x^3 satisfy all the above criterions.

Dhruv Somani - 4 years, 2 months ago

Problem changed.

Amogha Pokkulandra - 4 years, 2 months ago
Kushal Bose
Mar 19, 2017

p ( x ) = − ( x − 1 ) ( x − 2 ) ( x − 3 ) + x 3 p(x)=-(x-1)(x-2)(x-3) + x^3

The above equation satisfies all criterion and also a quadratic polynomial

So, p ( 4 ) = 58 p(4)=58

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