1. **State the problem:** Divide the polynomial $$9x^3 + 24x^2 + 18x + 14$$ by the binomial $$3x + 5$$ and find the quotient and remainder. 2. **Formula and method:** We use polynomial long division. The dividend is $$9x^3 + 24x^2 + 18x + 14$$ and the divisor is $$3x + 5$$. 3. **Step 1:** Divide the leading term of the dividend $$9x^3$$ by the leading term of the divisor $$3x$$: $$\frac{9x^3}{3x} = 3x^2$$ This is the first term of the quotient. 4. **Step 2:** Multiply the divisor by $$3x^2$$: $$(3x + 5)(3x^2) = 9x^3 + 15x^2$$ 5. **Step 3:** Subtract this from the dividend: $$(9x^3 + 24x^2 + 18x + 14) - (9x^3 + 15x^2) = 9x^2 + 18x + 14$$ 6. **Step 4:** Divide the new leading term $$9x^2$$ by $$3x$$: $$\frac{9x^2}{3x} = 3x$$ Add $$3x$$ to the quotient. 7. **Step 5:** Multiply the divisor by $$3x$$: $$(3x + 5)(3x) = 9x^2 + 15x$$ 8. **Step 6:** Subtract: $$(9x^2 + 18x + 14) - (9x^2 + 15x) = 3x + 14$$ 9. **Step 7:** Divide $$3x$$ by $$3x$$: $$\frac{3x}{3x} = 1$$ Add $$1$$ to the quotient. 10. **Step 8:** Multiply the divisor by $$1$$: $$(3x + 5)(1) = 3x + 5$$ 11. **Step 9:** Subtract: $$(3x + 14) - (3x + 5) = 9$$ 12. **Conclusion:** The quotient is $$3x^2 + 3x + 1$$ and the remainder is $$9$$. **Final answer:** Quotient: $$3x^2 + 3x + 1$$ Remainder: $$9$$