1. **State the problem:** We have a polynomial $p(x) = 2x^3 + ax^2 - 11x + b$. It is given that $p(x)$ is divisible by $(2x - 1)$ and when divided by $(x + 1)$ the remainder is 12. We need to find the values of $a$ and $b$. 2. **Use the Remainder Theorem:** - If $p(x)$ is divisible by $(2x - 1)$, then $p\left(\frac{1}{2}\right) = 0$. - The remainder when $p(x)$ is divided by $(x + 1)$ is $p(-1) = 12$. 3. **Set up equations:** Calculate $p\left(\frac{1}{2}\right)$: $$p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 - 11\left(\frac{1}{2}\right) + b = 0$$ 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: $$\cancel{4} \times \left(\frac{1}{4} + \frac{a}{4} - \frac{11}{2} + b\right) = \cancel{4} \times 0$$ $$1 + a - 22 + 4b = 0$$ Simplify: $$a + 4b - 21 = 0 \implies a + 4b = 21 \quad (1)$$ 4. Calculate $p(-1)$: $$p(-1) = 2(-1)^3 + a(-1)^2 - 11(-1) + b = 12$$ Simplify: $$2(-1) + a(1) + 11 + b = 12$$ $$-2 + a + 11 + b = 12$$ $$a + b + 9 = 12$$ $$a + b = 3 \quad (2)$$ 5. **Solve the system of equations:** From (1): $a + 4b = 21$ From (2): $a + b = 3$ Subtract (2) from (1): $$\cancel{a} + 4b - (\cancel{a} + b) = 21 - 3$$ $$4b - b = 18$$ $$3b = 18$$ $$b = 6$$ Substitute $b=6$ into (2): $$a + 6 = 3$$ $$a = 3 - 6 = -3$$ **Final answer:** $$a = -3, \quad b = 6$$