\(n^2+6n-27=\\
n^2-3n+9n-27=\\
n(n-3)+9(n-3)=\\
(n+9)(n-3)\)
For the above product to be a prime number, one of the factors must be a prime number and the other must be equal to 1.
\(n+9=1\\
n=-8\\\\
-8-3=-11\)
The first factor is equal 1 for \(n=-8\), but the other is euqal -11, which is not a prime number.
\(n-3=1\\
n=4\\\\
4+9=13\)
The second factor is equal 1 for \(n=4\) and the first factor is equal 13, which is a prime number.
So, \(n^2+6n-27\) is a prime number for \(n=4\)
