# Mock AIME 1 2007-2008 Problems/Problem 13

## Problem

Let $F(x)$ be a polynomial such that $F(6) = 15$ and $$\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}$$ for $x \in \mathbb{R}$ such that both sides are defined. Find $F(12)$.

## Solution

Combining denominators and simplifying, $$\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}$$ It becomes obvious that $F(x) = ax(x-1)$, for some constant $a$, matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that $$(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)$$

Since $3x$ and $3x-1$ divides the right side of the equation, $3x$ and $3x-1$ divides the left side of the equation. Thus $3x(3x-1)$ divides $F(3x)$, so $x(x-1)$ divides $F(x)$.

It is easy to see that $F(x)$ is a quadratic, thus $F(x)=ax(x-1)$ as desired.

By the given, $F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12$. Thus, $F(12) = \frac{1}{2}(12)(11) = \boxed{066}$.

## See also

 Mock AIME 1 2007-2008 (Problems, Source) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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