How to write $x^{p-1}-1$ as $(p-1)x^{p-2}\cdot g(x)+(p-1)\cdot r(x)$

Question

How to write $x^{p-1}-1$ as $(p-1)x^{p-2}\cdot g(x)+(p-1)\cdot r(x)$? How to find the appropriate polynomials $g(x),r(x)$

$\deg r(x)$ needs to be less than $p-2$

For context I am trying to show that if the caracteristic of the field over which those polynomials lie divides $p-1$ than $x^p-x$ has a multiple root.

I tried to work with $g(x)=x$ but it seems like it's not going to work this way...

Answer 1

Over a field of characteristic $\ell$ where $\ell\mid(p-1)$, then $$x^p-x=x(x^{(p-1)/\ell}-1)^\ell$$ and so certainly $1$ is zero with multiplicity at least $\ell$.