1. **State the problem:** We need to divide the cubic polynomial $$x^3 + 8x^2 + 19x + 12$$ by the linear polynomial $$x + 4$$ using algebraic long division to find constants $$A$$, $$B$$, and $$C$$ such that: $$\frac{x^3 + 8x^2 + 19x + 12}{x + 4} = A x^2 + B x + C$$ 2. **Recall the division process:** When dividing polynomials, we divide the highest degree term of the dividend by the highest degree term of the divisor, multiply the divisor by that result, subtract, and repeat with the remainder. 3. **Step 1:** Divide the leading term $$x^3$$ by $$x$$: $$\frac{x^3}{x} = x^2$$ So, $$A = 1$$. 4. **Step 2:** Multiply the divisor $$x + 4$$ by $$x^2$$: $$x^2(x + 4) = x^3 + 4x^2$$ 5. **Step 3:** Subtract this from the original polynomial: $$(x^3 + 8x^2 + 19x + 12) - (x^3 + 4x^2) = (8x^2 - 4x^2) + 19x + 12 = 4x^2 + 19x + 12$$ 6. **Step 4:** Divide the new leading term $$4x^2$$ by $$x$$: $$\frac{4x^2}{x} = 4x$$ So, $$B = 4$$. 7. **Step 5:** Multiply the divisor by $$4x$$: $$4x(x + 4) = 4x^2 + 16x$$ 8. **Step 6:** Subtract this from the current remainder: $$(4x^2 + 19x + 12) - (4x^2 + 16x) = (19x - 16x) + 12 = 3x + 12$$ 9. **Step 7:** Divide the leading term $$3x$$ by $$x$$: $$\frac{3x}{x} = 3$$ So, $$C = 3$$. 10. **Step 8:** Multiply the divisor by $$3$$: $$3(x + 4) = 3x + 12$$ 11. **Step 9:** Subtract this from the current remainder: $$(3x + 12) - (3x + 12) = 0$$ No remainder remains, confirming the division is exact. **Final answer:** $$\frac{x^3 + 8x^2 + 19x + 12}{x + 4} = x^2 + 4x + 3$$ Thus, $$A = 1$$, $$B = 4$$, and $$C = 3$$.