1. **State the problem:** Factor the quadratic expression $$5x^2 + 19x + 18$$. 2. **Recall the factoring formula:** For a quadratic $$ax^2 + bx + c$$, we look for two binomials $$(mx + n)(px + q)$$ such that: - $$m \times p = a$$ - $$n \times q = c$$ - $$m \times q + n \times p = b$$ 3. **Apply to our problem:** Here, $$a=5$$, $$b=19$$, and $$c=18$$. 4. **Find factors of $$a$$ and $$c$$:** - Factors of $$5$$: 5 and 1 - Factors of $$18$$: 1 and 18, 2 and 9, 3 and 6 5. **Try combinations:** - Try $$(5x + 2)(x + 9)$$: - Multiply outer and inner terms: $$5x \times 9 = 45x$$ and $$2 \times x = 2x$$ - Sum: $$45x + 2x = 47x$$ (not 19x) - Try $$(5x + 9)(x + 2)$$: - Outer and inner: $$5x \times 2 = 10x$$ and $$9 \times x = 9x$$ - Sum: $$10x + 9x = 19x$$ (matches $$b$$) 6. **Verify the product:** $$ (5x + 9)(x + 2) = 5x^2 + 10x + 9x + 18 = 5x^2 + 19x + 18 $$ 7. **Conclusion:** The factored form is $$\boxed{(5x + 9)(x + 2)}$$.