This problem is equivalent to showing that $(3x^5 + 5x^3 + 7x)$ is always a multiple of $15$. In order to show this, it is sufficient to show this expression is a multiple of both $3$ and $5$ for every $x$.

By Fermat little theorem, we have $x^3\equiv x\pmod{3}$ and $x^5\equiv x\pmod{5}$. Therefore $$\begin{array}{l} 3x^5 + 5x^3 + 7x \equiv 0+ 2x^3 + x\equiv 2x + x \equiv 3x\equiv 0 \pmod{3} \\ 3x^5 + 5x^3 + 7x \equiv 3x + 0 + 2x \equiv 5x\equiv 0\pmod{5}  \end{array}$$

Therefore, we conclude the claim holds.