Divisibility of polynomials 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.
Any suggestion?
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.
 A: Since $(f,f')\ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$. 
A: I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^{r-1}h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^{r-1}$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^{r-1}$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.

In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
