1. **State the problem:** Find values of $a$ and $b$ such that the polynomial $p(x) = 2x^3 + ax^2 - 11x + b$ is divisible by $(2x - 1)$ and leaves remainder 12 when divided by $(x + 1)$. 2. **Use the Remainder Theorem:** If $p(x)$ is divisible by $(2x - 1)$, then $p(\frac{1}{2}) = 0$. 3. **Apply divisibility condition:** $$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 - 11\left(\frac{1}{2}\right) + b = 0$$ Simplify: $$2\times \frac{1}{8} + a \times \frac{1}{4} - \frac{11}{2} + b = 0$$ $$\frac{1}{4} + \frac{a}{4} - \frac{11}{2} + b = 0$$ Multiply entire equation by 4 to clear denominators: $$1 + a - 22 + 4b = 0$$ Simplify: $$a + 4b - 21 = 0 \implies a + 4b = 21$$ 4. **Use remainder condition:** When divided by $(x + 1)$, remainder is 12, so $p(-1) = 12$. 5. **Apply remainder condition:** $$p(-1) = 2(-1)^3 + a(-1)^2 - 11(-1) + b = 12$$ Simplify: $$-2 + a + 11 + b = 12$$ $$a + b + 9 = 12$$ $$a + b = 3$$ 6. **Solve the system of equations:** $$\begin{cases} a + 4b = 21 \\ a + b = 3 \end{cases}$$ Subtract second from first: $$ (a + 4b) - (a + b) = 21 - 3 $$ $$ 3b = 18 $$ $$ b = 6 $$ 7. **Find $a$:** $$ a + b = 3 \implies a + 6 = 3 \implies a = 3 - 6 = -3 $$ **Final answer:** $$a = -3, \quad b = 6$$