Solving the Polynomial Equation ( frac{f(3x)}{f(x)} frac{729x-3}{x-243} ): A Comprehensive Guide

When solving polynomial equations, one crucial aspect is recognizing when such an equation holds for infinitely many values of ( x ). This indicates that the equation is actually true for all real values of ( x ). In this article, we will explore a detailed solution to the equation ( frac{f(3x)}{f(x)} frac{729x-3}{x-243} ) using a step-by-step approach.

Understanding the Equation

Given the polynomial equation:

[ frac{f(3x)}{f(x)} frac{729x-3}{x-243} ]

This equation must hold for infinitely many real values of ( x ) and, as a result, it must be true for all real values of ( x ). This is a non-trivial property that significantly simplifies the problem.

Initial Reductions

We start by cross-multiplying to eliminate the fractions:

[ x-243f(3x) 729 x-3f(x) ]

Two immediate reductions can be made by evaluating the equation at specific points:

Setting ( x 243 ):

[ 0 f(243) quad Rightarrow quad f(243) 0 ]

Setting ( x 3 ):

[ f(9) 0 ]

Divisibility and Factors

Given that ( f(243) 0 ) and ( f(9) 0 ), we can start factoring ( f(x) ) as:

[ f(x) (x-243)(x-9)g(x) ]

Substituting this factorization into the original equation gives:

[ frac{(3x-243)(3x-9)g(3x)}{(x-243)(x-9)g(x)} frac{729x-3}{x-243} ]

Canceling the common factors, we get:

[ frac{3x-243}{x-243} cdot frac{3x-9}{x-9} cdot frac{g(3x)}{g(x)} frac{729x-3}{x-243} ]

Focusing on the remaining factors, we have:

[ frac{3x-9}{x-9} 729 cdot frac{1}{3x-243} ]

Finding the Form of ( g(x) )

By further simplifying and recognizing the pattern, we find:

Setting ( x 81 ):

[ g(81) 0 ]

Setting ( x 27 ):

[ g(27) 0 ]

Thus, we can factor ( g(x) ) as:

[ g(x) (x-27)(x-81)h(x) ]

Substituting this back, we get:

[ f(x) (x-243)(x-9)(x-27)(x-81)h(x) ]

Substituting into the original equation:

[ frac{(3x-243)(3x-9)(3x-27)(3x-81)h(3x)}{(x-243)(x-9)(x-27)(x-81)h(x)} frac{729x-3}{x-243} ]

Canceling the common factors, we get:

[ frac{3x-243}{x-243} cdot frac{3x-9}{x-9} cdot frac{3x-27}{x-27} cdot frac{3x-81}{x-81} cdot frac{h(3x)}{h(x)} frac{729x-3}{x-243} ]

Further simplification shows:

[ frac{3x-9}{x-9} cdot frac{3x-27}{x-27} cdot frac{3x-81}{x-81} 729 ]

Thus, we find:

Setting ( x 3 ):

[ frac{h(3x)}{h(x)} 3 ]

This implies:

[ h(x) cx^2 ]

Where ( c ) is a non-zero constant. Therefore, the polynomial ( f(x) ) is:

[ f(x) c(x-243)(x-9)(x-27)(x-81)x^2 ]

Conclusion

The polynomial ( f(x) c(x-243)(x-9)(x-27)(x-81)x^2 ) satisfies the given equation for any non-zero constant ( c ). This solution is derived from the constraints imposed by the initial reductions and the factorization process, demonstrating the power of algebraic manipulation in solving complex polynomial equations.