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? 

Hint: $\;6x^4+50 = 0\;$ has $4$ complex roots, and for each one $\;|x_j| = \sqrt[4]{50/6}=\sqrt{5}/\sqrt[4]{3}\;$.
If $3x^2+4x+m$ is a factor of $6x^4+50$ then its two complex roots will be among $x_j \mid j=1 \dots 4$. By Vieta's formulas their product $x_j x_k=m/3 \implies |x_j x_k|=5 / \sqrt{3} = |m/3|$. But $5 / \sqrt{3}$ is irrational, so there exists no $m \in \mathbb{Z}$ such that $5 / \sqrt{3} = |m/3|$.
