1. **Problem 1**: Given a polynomial $P(x) = 3x^3 + ax^2 + bx - 9$, it leaves a remainder of $-7$ when divided by $x+1$ and a remainder of $23$ when divided by $x-2$. We need to find the remainder when $P(x)$ is divided by $2x-1$. 2. When divided by $x+1$, the remainder is $P(-1) = -7$. 3. When divided by $x-2$, the remainder is $P(2) = 23$. 4. We write these equations: $$P(-1) = 3(-1)^3 + a(-1)^2 + b(-1) - 9 = -3 + a - b - 9 = a - b - 12 = -7$$ From this, we get: $$a - b = 5$$ 5. Similarly for $x=2$: $$P(2) = 3(2)^3 + a(2)^2 + b(2) - 9 = 24 + 4a + 2b - 9 = 15 + 4a + 2b = 23$$ From this, we get: $$4a + 2b = 8$$ 6. Solve the system: Using $a - b = 5$, so $b = a - 5$. Substitute into $4a + 2b = 8$: $$4a + 2(a - 5) = 8 \\ 4a + 2a - 10 = 8 \\ 6a = 18 \\ a = 3$$ Then $b = 3 - 5 = -2$. 7. The polynomial is now: $$P(x) = 3x^3 + 3x^2 - 2x - 9$$ 8. Find the remainder when $P(x)$ is divided by $2x - 1$. The divisor $2x - 1 = 0$ implies $x = \frac{1}{2}$. 9. Evaluate $P(\frac{1}{2})$: $$P(\frac{1}{2}) = 3(\frac{1}{2})^3 + 3(\frac{1}{2})^2 - 2(\frac{1}{2}) - 9 = 3(\frac{1}{8}) + 3(\frac{1}{4}) - 1 - 9 = \frac{3}{8} + \frac{3}{4} - 10 = \frac{3}{8} + \frac{6}{8} - 10 = \frac{9}{8} - 10 = -\frac{71}{8}$$ 10. Therefore, the remainder when $P(x)$ is divided by $2x - 1$ is $-\frac{71}{8}$. --- 11. **Problem 2**: Is $(x+1)$ a factor of $x^5 - 1$? 12. To check if $(x+1)$ is a factor, evaluate the polynomial at $x = -1$: $$(-1)^5 - 1 = -1 - 1 = -2 \ne 0$$ 13. Since $P(-1) \ne 0$, $(x+1)$ is NOT a factor. 14. **Answer: FALSE**. **Final answers:** - The remainder when $P(x)$ is divided by $2x-1$ is $-\frac{71}{8}$. - $(x+1)$ is NOT a factor of $x^5 - 1$.