Prove there is no integer $m$ such that $3x^2+4x+m$ is a factor of $6x^4+50$ in $\mathbb{Z}[x]$. Prove there is no integer $m$ such that $3x^2+4x+m$ is a factor of $6x^4+50$ in $\mathbb{Z}[x]$.
I understand that I will need to use the division algorithm on polynomials, however I am confused why $m$ excludes $3x^2+4x+m$ as a factor. I found the division of these polynomials will yield fractional coefficients which are not in $\mathbb{Z}[x]$, thus no integer $m$ will make $3x^2+4x+m$ a factor. Is this the correct train of thought? 

$ \begin{align}{\bf Hint}\quad  {\rm mod}\ 3\!:\ \  & (3x^2\!+\!4x\!+\!m)\,(2x^2\!+\!bx\!+\!c)\\ 
&\equiv\ \ \ \ \, (x+m)\,(2x^2+\cdots)\\ 
& \equiv  \  \ 2\,x^{\large\color{#c00}3}+\,\cdots  \end{align} $
$\, 6x^4\!+50\,\equiv 50\ $ has degree $< \color{#c00}3\ $ so they cannot be equal.
