1. **State the problem:** Factor the quadratic expression $$y^2 - 9y + 18$$. 2. **Recall the factoring formula:** For a quadratic of the form $$y^2 + by + c$$ where the leading coefficient is 1, we look for two numbers that multiply to $$c$$ and add to $$b$$. 3. **Identify values:** Here, $$b = -9$$ and $$c = 18$$. 4. **Find two numbers:** We need two numbers that multiply to $$18$$ and add to $$-9$$. These numbers are $$-6$$ and $$-3$$ because $$-6 \times -3 = 18$$ and $$-6 + (-3) = -9$$. 5. **Write the factored form:** Using these numbers, the factorization is: $$y^2 - 9y + 18 = (y - 6)(y - 3)$$ 6. **Check by expansion:** Expanding: $$(y - 6)(y - 3) = y^2 - 3y - 6y + 18 = y^2 - 9y + 18$$ This matches the original expression, confirming the factorization is correct. **Final answer:** $$\boxed{(y - 6)(y - 3)}$$