Question

Find two rational functions f(x) and g(x) such that $f(x) * g(x) = x^{2}$ and $\frac{f(x)}{g(x)} = \frac{(x-1)^2}{(x+2)^2}$

Answer 1

$\dfrac fg=\dfrac{(x-1)^2}{(x+2)^2}$
$\implies f=g\dfrac{(x-1)^2}{(x+2)^2}$
$\implies f^2=fg\dfrac{(x-1)^2}{(x+2)^2}$
$\implies f^2=x^2\dfrac{(x-1)^2}{(x+2)^2}$
$\implies\sqrt{f^2}=\sqrt{x^2\dfrac{(x-1)^2}{(x+2)^2}}$
$\implies|f|=\left|\dfrac{x(x-1)}{x+2}\right|$

Similarly, we can show that

$|g|=\left|\dfrac{x(x+2)}{x-1}\right|$

Then from here we can solve for $f$ or $g$ by taking either to be both positive, or both negative. That is,

$f(x)=\dfrac{x(x-1)}{x+2},\,g(x)=\dfrac{x(x+2)}{x-1}$

or

$f(x)=-\dfrac{x(x-1)}{x+2},\,g(x)=-\dfrac{x(x+2)}{x-1}$

(Note that this is only true for $x\neq1$, $x\neq-2$, and $x\neq0$.)