Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

## Theory of Equations – AIME 2015

Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

- is 107
- is 72
- is 840
- cannot be determined from the given information

**Key Concepts**

**S**eries

Theory of Equations

Algebra

## Check the Answer

But try the problem first…

Answer: is 72.

AIME, 2015, Question 10.

Polynomials by Barbeau.

## Try with Hints

First hint

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Second Hint

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Final Step

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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