1. **State the problem:** We have a polynomial $g(x)$ and know the remainders when divided by two linear polynomials: - When divided by $x+2$, remainder is $-1$. - When divided by $2x-1$, remainder is $12$. We want to find the remainder when $g(x)$ is divided by the product $(x+2)(2x-1)$. 2. **Recall the Remainder Theorem:** - The remainder when dividing by a linear polynomial $x-a$ is $g(a)$. - When dividing by a quadratic polynomial like $(x+2)(2x-1)$, the remainder is a polynomial of degree less than 2, i.e., a linear polynomial $R(x) = Ax + B$. 3. **Set up the remainder polynomial:** Let the remainder be $R(x) = Ax + B$. 4. **Use the given conditions:** - Since dividing by $x+2$ leaves remainder $-1$, then $g(-2) = R(-2) = A(-2) + B = -2A + B = -1$. - Since dividing by $2x-1$ leaves remainder $12$, then $g(\frac{1}{2}) = R(\frac{1}{2}) = A(\frac{1}{2}) + B = \frac{A}{2} + B = 12$. 5. **Solve the system of equations:** $$\begin{cases} -2A + B = -1 \\ \frac{A}{2} + B = 12 \end{cases}$$ 6. **Subtract the first equation from the second:** Multiply the second equation by 2 to clear fraction: $$A + 2B = 24$$ Now system is: $$\begin{cases} -2A + B = -1 \\ A + 2B = 24 \end{cases}$$ 7. **Solve for $A$ and $B$:** Multiply first equation by 2: $$-4A + 2B = -2$$ Add to second equation: $$(-4A + 2B) + (A + 2B) = -2 + 24$$ $$-3A + 4B = 22$$ From first equation: $$B = -1 + 2A$$ Substitute into above: $$-3A + 4(-1 + 2A) = 22$$ $$-3A -4 + 8A = 22$$ $$5A - 4 = 22$$ $$5A = 26$$ $$A = \frac{26}{5}$$ Then: $$B = -1 + 2 \times \frac{26}{5} = -1 + \frac{52}{5} = \frac{-5 + 52}{5} = \frac{47}{5}$$ 8. **Write the remainder polynomial:** $$R(x) = \frac{26}{5}x + \frac{47}{5}$$ **Final answer:** The remainder when $g(x)$ is divided by $(x+2)(2x-1)$ is $$\boxed{\frac{26}{5}x + \frac{47}{5}}$$.