Finding Primes

Because the question allows for "real values of $n$ ", and the image of the expression is all real numbers, hence it follows that all primes can be expressed as $2n^5-20n^4+70n^3-100n^2+48n+5$ for some real number $n$ . Thus, the answer is "more than 6".

Note: The solution below is written up for the case where $n$ is an integer.

$\large p = 2n^5-20n^4+70n^3-100n^2+48n+5$ $\large p = 2n(n^4 -10n^3 +35n^2 -50n +24) + 5$ $\large p = 2n(n-1)(n-2)(n-3)(n-4) + 5$

$\large p = \begin{cases} 5 \quad \quad \quad ; n = 0 \\ 5 \quad \quad \quad ; n = 1 \\ 5 \quad \quad \quad ; n = 2 \\ 5 \quad \quad \quad ; n = 3 \\ 5 \quad \quad \quad ; n = 4 \\ 125 \quad \quad ; n = 5 \quad \quad \color{#3D99F6}{\text{multiple of 5}} \\ 1545 \ \quad ; n = 6 \quad \quad \color{#3D99F6}{\text{multiple of 5}} \end{cases}$

Hence, p is a prime number for $n = 0,1,2,3$ and $4$ .