To factor the trinomial \( 15n^2 - 27n - 6 \) completely, we start by looking for two numbers that multiply to the product of the leading coefficient (15) and the constant term (-6), which is \( 15 \times -6 = -90 \), and also add up to the middle coefficient (-27).
We need to find two numbers that meet these criteria. After checking possible pairs, we find that the numbers \(-30\) and \(3\) work since:
\[ -30 \times 3 = -90 \] \[ -30 + 3 = -27 \]
Next, we can use these two numbers to break down the middle term \(-27n\):
\[ 15n^2 - 30n + 3n - 6 \]
Now, we group the terms:
\[ (15n^2 - 30n) + (3n - 6) \]
Now, we factor each group:
- In the first group \(15n^2 - 30n\), we can factor out \(15n\):
\[ 15n(n - 2) \]
- In the second group \(3n - 6\), we can factor out \(3\):
\[ 3(n - 2) \]
Now we combine these factored groups:
\[ 15n(n - 2) + 3(n - 2) \]
We can see that \((n - 2)\) is a common factor:
\[ (n - 2)(15n + 3) \]
Finally, we notice that \(15n + 3\) can be factored further by pulling out the 3:
\[ 3(5n + 1) \]
Thus, we rewrite the expression:
\[ (n - 2)(3)(5n + 1) = 3(n - 2)(5n + 1) \]
Therefore, the completely factored form of the trinomial \(15n^2 - 27n - 6\) is:
\[ \boxed{3(n - 2)(5n + 1)} \]