GCD identity for univariate polynomials I would like to know if my solution to the following question concerning the gcd of two polynomial correct> 
Question:  
Suppose $f(x)$, $g(x)$ and $h(x)$ are polynomials in $F[x]$.  Let $gcd(f(x),g(x)h(x))=1$.  Show that $gcd(f(x),g(x))=gcd(f(x),h(x))=1$
Proof:  Since $gcd(f(x),g(x)h(x))=1$, then there exists some polynomials $u(x)$ and $v(x)$ in $F[x]$ where  $f(x)u(x)+(g(x)h(x))v(x)=1$. $(*)$ 
Applying associativity of multiplication in $(*)$ gives $(g(x)h(x))v(x)=g(x)(h(x)v(x))$ then $f(x)u(x)+(g(x)h(x))v(x)=f(x)u(x)+g(x)(h(x)v(x))=1$ which implies $gcd(f(x), g(x)=1$.
Applying commutativity to $(*)$, then again associativity of multiplication to $(*)$ gives $(g(x)h(x))v(x)=(h(x)g(x))v(x))$ and $(h(x)g(x))v(x))=h(x)(g(x)v(x))$ in $(*)$ implies that $f(x)u(x)+(g(x)h(x))v(x)=f(x)u(x)+(h(x)g(x))v(x)=f(x)u(x)+h(x)(g(x)v(x))=1$, which gives $gcd(f(x), h(x))=1$ also.  Hence we have $gcd(f(x), g(x))=gcd(f(x), h(x))=1$
Thank you in advance
 A: The gcd $(a,b)\,$ is characterized by the following fundamental gcd Universal Property
$$ c\mid a,b\,\color{#0a0}\Leftarrow\!\color{#c00}\Rightarrow\, c\mid (a,b)\quad \text{[gcd Universal Property]}\qquad$$
For $\,c = (a,b)\,$ arrow $\color{#0a0}{(\Leftarrow)}$ shows $\,(a,b)\mid a,b,\,$ i.e. $\,(a,b)\,$ is a common divisor of $\,a,b,\,$ and the opposite arrow $\color{#c00}{(\Rightarrow)}$ shows that $\,(a,b)\,$ is divisible by every common divisor of $\,a,b,\,$ so $\,(a,b)\,$ is a greatest (degree) common divisor  (unique if we normalize it to be monic, i.e. lead coef $=1).\,$
Therefore generally $\,(f,g)\mid f,g\,\Rightarrow\, (f,g)\mid f,gh\,\color{#c00}\Rightarrow\, \color{#90f}{(f,g)}\mid \color{#0a0}{(f,gh)}$.
Further, in your particular case we are given that $\,\color{#0a0}{(f,gh)\!=1}\,$ thus  $\color{#90f}{(f,g)}\! =\! 1$
Your use of the Bezout identity to deduce this is essentially repeating the Bezout-based proof of the gcd Universal Property in this particular case. Instead you should abstract out and prove this basic property (whose proof is easier), then invoke it as above (by name). This will work more generally in rings where there is no Bezout identity (e.g. $\,\Bbb Z[x]\,$ or $\,\Bbb Q[x,y]\:\!)\,$ but the gcd universal property still holds true (it is the definition of the gcd in general domains).
