1. **State the problem:** Factor the trinomial $$15n^2 - 27n - 6$$ completely. 2. **Recall the factoring formula:** For a quadratic trinomial $$ax^2 + bx + c$$, we look for two numbers that multiply to $$a \times c$$ and add to $$b$$. 3. **Calculate the product and sum:** Here, $$a = 15$$, $$b = -27$$, and $$c = -6$$. Calculate $$a \times c = 15 \times (-6) = -90$$. We need two numbers that multiply to $$-90$$ and add to $$-27$$. 4. **Find the pair:** The numbers $$-30$$ and $$3$$ satisfy this because $$-30 \times 3 = -90$$ and $$-30 + 3 = -27$$. 5. **Rewrite the middle term:** $$15n^2 - 30n + 3n - 6$$ 6. **Group terms:** $$(15n^2 - 30n) + (3n - 6)$$ 7. **Factor each group:** $$15n(n - 2) + 3(n - 2)$$ 8. **Factor out the common binomial:** $$(15n + 3)(n - 2)$$ 9. **Simplify the first factor:** Factor out 3: $$3\cancel{(5n + 1)}\cancel{(n - 2)}$$ Actually, factor out 3: $$3(5n + 1)(n - 2)$$ 10. **Final factored form:** $$3(5n + 1)(n - 2)$$ **Answer:** $$15n^2 - 27n - 6 = 3(5n + 1)(n - 2)$$