Question

Factor the following trinomial completely (do not solve): 15n2−27n−6

Answer 1

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:

1. In the first group \(15n^2 - 30n\), we can factor out \(15n\):

\[
15n(n - 2)
\]

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:

\[
\boxed{3(n - 2)(5n + 1)}
\]