Factoring $p^3x-pc^2x^{-1}+c(1-p^4)=0$ to $(x-cp)(p^3x+c)=0$ without trial and error

Question

I was doing a AS Further Pure Past Paper from 2018 from the Edexcel spec. For question 5, which was about rectangular hyperbola, I needed to factorize this: $$ p^{3}x - pc^{2}x^{-1}+c(1-p^{4})=0 \tag1 $$ in order to find the value of $x$ in terms of $p$ and $c$. I know that the answer is: $$ (x-cp)(p^{3}x+c)=0 \tag2 $$ but I don't know how to reach that answer without trial and error.

I was hoping that someone could outline a method to solve for $x$.

Answer 1

since the normal meets the curve at $(cp,\frac{c}{p})$, $(x-cp)$ must be a factor

Answer 2

$p^3x-pc^2x^{-1}+c(1-p^4)=0$

$\implies x^{-1} (p^3x^2 - pc^2 + cx - cp^4x) = 0$

$\implies (p^3x^2 - pc^2 + cx - cp^4x) = 0$

$\implies ((p^3x^2 + cx) - (pc^2 + cp^4x)) = 0$

$\implies (x(p^3x + c) - pc(c + p^3x + c)) = 0$

$\implies (x - cp)(p^3x + c) = 0$