Question

Factor the following trinomial completely (do not solve)
15n^2 - 27n - 6

Answer 1

To factor the trinomial \(15n^2 - 27n - 6\) completely, we can follow these steps:

1. **Identify a, b, and c** from the standard quadratic form \(ax^2 + bx + c\):
\[
a = 15, \quad b = -27, \quad c = -6
\]

2. **Multiply \(a\) and \(c\)**:
\[
ac = 15 \times -6 = -90
\]

3. **Find two numbers that multiply to \(ac\) (-90) and add to \(b\) (-27)**. The numbers we are looking for are -30 and 3 because:
\[
-30 \times 3 = -90 \quad \text{and} \quad -30 + 3 = -27
\]

4. **Rewrite the middle term using these two numbers**:
\[
15n^2 - 30n + 3n - 6
\]

5. **Group the terms**:
\[
(15n^2 - 30n) + (3n - 6)
\]

6. **Factor out the common factors in each group**:
\[
15n(n - 2) + 3(n - 2)
\]

7. **Factor out the common binomial factor**:
\[
(n - 2)(15n + 3)
\]

8. **Finally, to express it in its completely factored form, we can also factor out the 3 from the second binomial**:
\[
(n - 2)(3(5n + 1)) = 3(n - 2)(5n + 1)
\]

Thus, the complete factorization of the trinomial \(15n^2 - 27n - 6\) is:
\[
\boxed{3(n - 2)(5n + 1)}
\]