1. **Problem Statement:** We have a polynomial $g(x)$ such that when divided by $(x+1)$, $(x-1)$, and $(x-2)$, the remainders are 1, $\frac{1}{2}$, and $\frac{2}{3}$ respectively. We define $Q(x) = (x+3)g(x) - 2$ and need to show that $Q(x)$ is divisible by $(x+1)$, $(x-1)$, and $(x-2)$. 2. **Key Formula and Rules:** - The Remainder Theorem states that the remainder when a polynomial $f(x)$ is divided by $(x-a)$ is $f(a)$. - If $Q(x)$ is divisible by $(x-a)$, then $Q(a) = 0$. 3. **Using the Remainder Theorem on $g(x)$:** - Since remainder when $g(x)$ is divided by $(x+1)$ is 1, we have $g(-1) = 1$. - Since remainder when $g(x)$ is divided by $(x-1)$ is $\frac{1}{2}$, we have $g(1) = \frac{1}{2}$. - Since remainder when $g(x)$ is divided by $(x-2)$ is $\frac{2}{3}$, we have $g(2) = \frac{2}{3}$. 4. **Evaluate $Q(x)$ at the roots of divisors:** - Calculate $Q(-1) = (-1+3)g(-1) - 2 = 2 \times 1 - 2 = 0$. - Calculate $Q(1) = (1+3)g(1) - 2 = 4 \times \frac{1}{2} - 2 = 2 - 2 = 0$. - Calculate $Q(2) = (2+3)g(2) - 2 = 5 \times \frac{2}{3} - 2 = \frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3} \neq 0$. 5. **Re-examining the last step:** The problem states $Q(x)$ is divisible by $(x+1)$, $(x-1)$, and $(x-2)$, so $Q(2)$ must be zero. Check calculations: $Q(2) = (2+3)g(2) - 2 = 5 \times \frac{2}{3} - 2 = \frac{10}{3} - 2 = \frac{10}{3} - \frac{6}{3} = \frac{4}{3}$ which is not zero. This suggests a possible typo or misunderstanding in the problem statement or remainder values. 6. **Assuming the problem wants to show divisibility by $(x+1)$ and $(x-1)$ only:** Since $Q(-1) = 0$ and $Q(1) = 0$, $Q(x)$ is divisible by $(x+1)$ and $(x-1)$. 7. **Conclusion:** $Q(x)$ is divisible by $(x+1)$ and $(x-1)$ as shown by $Q(-1) = 0$ and $Q(1) = 0$. If the problem requires divisibility by $(x-2)$ as well, the given remainder values or $Q(x)$ definition may need revision. **Final answer:** $$Q(-1) = 0, \quad Q(1) = 0, \quad Q(2) \neq 0,$$ so $Q(x)$ is divisible by $(x+1)$ and $(x-1)$ but not by $(x-2)$ with the given data.