Q:

Show that if $1$ is a gcd of $r(x)$ and $s(x)$, and $r(x)$ is a divisor of $s(x)t(x)$, then $r(x)$ is a divisor of $t(x)$.

$F$ is a field and $r(x),t(x),s(x)\in F[x]$. 
Show that if $1$ is a gcd of $r(x)$ and $s(x)$, and $r(x)$ is a divisor of $s(x)t(x)$, then $r(x)$ is a divisor of $t(x)$.
My attempt:
I'm having trouble figuring out what it means if the gcd of two polynomials under a field is 1. 
I first tried to make an analogy of what this means with numbers in $\mathbb Z$, stating that these polynomials are "coprime". 
I know that using this analogy, a "prime" polynomial is irreducible. So does this mean that $s(x)$ cannot be divided by $r(x)$ and $r(x)$ cannot be divided by $s(x)$?
By this, I concluded that since $r(x)$ cannot divide $s(x)$, but it divides $s(x)t(x)$, it must be that $r(x)$ divides $t(x)$.

A:

Hint $\ $ By Bezout, $\,1 = (r,s)\,\Rightarrow\, 1 = ar+bs\,$ so $\, r\mid \color{#c00}{st}\,\Rightarrow\ r\mid a\color{#c00}rt,b\color{#c00}{st}\,\Rightarrow\,r\mid(a\color{#c00}r+b\color{#c00}s)\color{#c00}t = t$
Remark $\ $ Just as for integers, the Euclidean algorithm for polynomials yields the Bezout Identity, for example see this answer. 

