$\gcd$ of polynomials over a field I have the polynomials $f,g\neq 0 $ over a field $F$. We know also that $\gcd(f,g)=1$ and 
$$
       \det \begin{pmatrix}
        a & b \\
        c & d \\
        \end{pmatrix}\neq 0.
$$
I need to prove that $\gcd(af+bg,cf+dg) = 1 $ for every $a,b,c,d \in F$. 
I really do not know how to start answer the question. Thanks for helpers!

HINT: What if $\gcd(af+bg,cf+dg) \neq 1$? 
Well, By using the Unique Factorization Property of the polynomial ring over a field you can find a prime polynomial $p$ such that $p \mid \gcd(af+bg,cf+dg)$. 
Does it give you any idea about how you should go ahead with this argument to get a contradiction?
2nd hint:
Well, since $p \mid af+bg$ and $p \mid cf+dg$ we can conclude that $p \mid d(af+bg)-b(cf+dg) \implies p \mid (ad-bc)f$ .
But $p$ is prime, and $ad-bc \neq 0$, therefore $\gcd(p, ad-bc) =1$ since $p$ is assumed to be a monic polynomial. so, you get $p \mid f$. the same reasoning gives you $p \mid g$. That means $p \mid \gcd(f,g)$ which gives you the contradiction.
