I am having trouble trying to prove the following statement:
Let $f = (f,f').g$ with $(f,f')\not= 1$. Both $f$ and $g$ are polynomials with rational coefficients. Then, $(f,f')$ and $g$ share roots.
My idea goes like this:
I name $d = (f,f')$. Since $d \not= 1$ then $f$ is not irreducible, so it may be written as $f(x) = p^2(x) \, . a(x)$ for some polynomial $a$ where $p$ is an irreducible factor of $d$.
Then $f'= 2\,p\,p' \, a + p^2 \, a'$ = $p \, (2 \, p' \, a + p \, a')$. This means that $p|f'$.
On the other hand, $f' = d' \, g + d \, g'$, ant since $p$ divides both $d$ and $f$, it must divide $d' g$.
So I believe that I need to show that $(p,d') = 1$ because in that case I can assure $p$ divides $g$, so $g$ and $f$ will share roots, since $p$ will divide both of them; but I am missing this final step.
EDIT: I corrected my question. I originally wrote "$f$ and $g$ share roots" instead of "$(f,f')$ and $g$ share roots" which is what I was really looking for.