1. **State the problem:** Find the values of $a$ and $b$ for the cubic polynomial $f(x) = 3x^3 + ax^2 + x + b$ given that $x + 1$ is a factor of $f(x)$ and when $f(x)$ is divided by $x - 4$, the remainder is $9b + 116$. 2. **Use the factor theorem:** Since $x + 1$ is a factor, $f(-1) = 0$. 3. **Evaluate $f(-1)$:** $$f(-1) = 3(-1)^3 + a(-1)^2 + (-1) + b = -3 + a - 1 + b = a + b - 4$$ Set equal to zero: $$a + b - 4 = 0$$ $$\Rightarrow a + b = 4$$ 4. **Use the remainder theorem:** The remainder when dividing by $x - 4$ is $f(4) = 9b + 116$. 5. **Evaluate $f(4)$:** $$f(4) = 3(4)^3 + a(4)^2 + 4 + b = 3 \times 64 + 16a + 4 + b = 192 + 16a + 4 + b = 196 + 16a + b$$ Set equal to remainder: $$196 + 16a + b = 9b + 116$$ 6. **Simplify the equation:** $$196 + 16a + b = 9b + 116$$ $$196 + 16a + b - 9b - 116 = 0$$ $$16a - 8b + 80 = 0$$ 7. **Rewrite:** $$16a - 8b = -80$$ Divide both sides by 8: $$\cancel{8} \times 2a - \cancel{8} b = \cancel{8} \times (-10)$$ $$2a - b = -10$$ 8. **Solve the system:** From step 3: $a + b = 4$ From step 7: $2a - b = -10$ Add the two equations: $$(a + b) + (2a - b) = 4 + (-10)$$ $$3a = -6$$ $$a = -2$$ Substitute $a = -2$ into $a + b = 4$: $$-2 + b = 4$$ $$b = 6$$ **Final values:** $$a = -2, \quad b = 6$$