# Math Help - f(x)

1. ## f(x)

Determine all polynomials $f(x)$

satisfying

$(x-243) f(3x) = 243 (x-1)f(x)$

for all $x$.

2. When $x=1$ the RHS = 0 so the LHS must be 0 as well; thus $f(3)=0.$

Substituting $x=3$ then gives RHS = 0 again, so LHS must be 0 $\implies\,f(9)=0.$

Continuing this way, we get $f(3)=f(9)=f(27)=f(81)=f(243)=0.$ Hence $f(x)$ must have these five roots and so we have

$f(x)\ =\ (x-3)(x-9)(x-27)(x-81)(x-243)g(x)$

where $g(x)$ is some polynomial.

$\therefore\ f(3x)\,=\,(3x-3)(3x-9)(3x-27)(3x-81)(3x-243)g(3x)$

$=\,243(x-1)(x-3)(x-9)(x-27)(x-81)g(3x)$

$\implies\ (x-243)f(3x)\,=\,243(x-1)(x-3)(x-9)(x-27)(x-81)(x-243)g(3x)$ $=$ $243(x-1)f(x)$

Thus the polynomial $g(x)$ must satisfy $g(x)=g(3x)$ for all $x$ and so must be a constant function. Hence

$\fbox{f(x)~=~a(x-3)(x-9)(x-27)(x-81)(x-243)}$

where $a$ is a real constant.