1. Problem 1: Given $p(x) = ax^3 + 5x^2 - 4x + b$, $(x+2)$ is a factor and remainder when divided by $(x+1)$ is 2. 2. Use factor theorem: $p(-2) = 0$ and remainder theorem: $p(-1) = 2$. 3. Calculate $p(-2) = a(-2)^3 + 5(-2)^2 - 4(-2) + b = -8a + 20 + 8 + b = -8a + b + 28 = 0$. 4. Calculate $p(-1) = a(-1)^3 + 5(-1)^2 - 4(-1) + b = -a + 5 + 4 + b = -a + b + 9 = 2$. 5. Simplify $-a + b + 9 = 2$ to $-a + b = -7$. 6. System of equations: $$\begin{cases} -8a + b + 28 = 0 \\ -a + b = -7 \end{cases}$$ 7. Subtract second from first: $(-8a + b + 28) - (-a + b) = 0 - (-7)$ gives $-7a + 28 = 7$. 8. Simplify: $-7a = -21$ so $a = 3$. 9. Substitute $a=3$ into $-a + b = -7$: $-3 + b = -7$ so $b = -4$. 10. Problem 2: $p(x) = ax^3 + x^2 + bx + 3$, divisible by $(2x - 1)$ and remainder when divided by $(x + 2)$ is 5. 11. Divisible by $(2x - 1)$ means $p(\frac{1}{2}) = 0$. 12. Remainder theorem: $p(-2) = 5$. 13. Calculate $p(\frac{1}{2}) = a(\frac{1}{2})^3 + (\frac{1}{2})^2 + b(\frac{1}{2}) + 3 = a\frac{1}{8} + \frac{1}{4} + \frac{b}{2} + 3 = 0$. 14. Multiply by 8: $a + 2 + 4b + 24 = 0$ so $a + 4b + 26 = 0$. 15. Calculate $p(-2) = a(-2)^3 + (-2)^2 + b(-2) + 3 = -8a + 4 - 2b + 3 = -8a - 2b + 7 = 5$. 16. Simplify: $-8a - 2b = -2$. 17. System: $$\begin{cases} a + 4b + 26 = 0 \\ -8a - 2b = -2 \end{cases}$$ 18. Multiply first by 2: $2a + 8b + 52 = 0$. 19. Add to second: $(-8a - 2b) + (2a + 8b + 52) = -2 + 0$ gives $-6a + 6b + 52 = -2$. 20. Simplify: $-6a + 6b = -54$ or $-a + b = -9$. 21. From first: $a = -4b - 26$. 22. Substitute into $-a + b = -9$: $-(-4b - 26) + b = -9$ gives $4b + 26 + b = -9$. 23. Simplify: $5b + 26 = -9$ so $5b = -35$ and $b = -7$. 24. Substitute $b = -7$ into $a = -4b - 26$: $a = -4(-7) - 26 = 28 - 26 = 2$. 25. Problem 3: $p(x) = 2x^3 + ax^2 + bx + 6$, remainder when divided by $(x+2)$ is -38 and by $(2x - 1)$ is 4. 26. $p(-2) = -38$ and $p(\frac{1}{2}) = 4$. 27. Calculate $p(-2) = 2(-2)^3 + a(-2)^2 + b(-2) + 6 = -16 + 4a - 2b + 6 = 4a - 2b - 10 = -38$. 28. Simplify: $4a - 2b = -28$. 29. Calculate $p(\frac{1}{2}) = 2(\frac{1}{2})^3 + a(\frac{1}{2})^2 + b(\frac{1}{2}) + 6 = 2\frac{1}{8} + a\frac{1}{4} + \frac{b}{2} + 6 = \frac{1}{4} + \frac{a}{4} + \frac{b}{2} + 6 = 4$. 30. Multiply by 4: $1 + a + 2b + 24 = 16$. 31. Simplify: $a + 2b + 25 = 16$ so $a + 2b = -9$. 32. System: $$\begin{cases} 4a - 2b = -28 \\ a + 2b = -9 \end{cases}$$ 33. Add equations: $(4a - 2b) + (a + 2b) = -28 + (-9)$ gives $5a = -37$. 34. So $a = -\frac{37}{5}$. 35. Substitute into $a + 2b = -9$: $-\frac{37}{5} + 2b = -9$. 36. Multiply by 5: $-37 + 10b = -45$. 37. Simplify: $10b = -8$ so $b = -\frac{4}{5}$. 38. Problem 4: $p(x) = ax^3 - 10x^2 + bx + 8$, $(x-2)$ is a factor of $p(x)$ and $p'(x)$. 39. Since $(x-2)$ is a factor of $p(x)$, $p(2) = 0$. 40. Since $(x-2)$ is a factor of $p'(x)$, $p'(2) = 0$. 41. Calculate $p(2) = a(2)^3 - 10(2)^2 + b(2) + 8 = 8a - 40 + 2b + 8 = 8a + 2b - 32 = 0$. 42. Calculate derivative $p'(x) = 3ax^2 - 20x + b$. 43. Calculate $p'(2) = 3a(2)^2 - 20(2) + b = 12a - 40 + b = 0$. 44. System: $$\begin{cases} 8a + 2b = 32 \\ 12a + b = 40 \end{cases}$$ 45. Multiply second by 2: $24a + 2b = 80$. 46. Subtract first from this: $(24a + 2b) - (8a + 2b) = 80 - 32$ gives $16a = 48$. 47. So $a = 3$. 48. Substitute into $12a + b = 40$: $12(3) + b = 40$ so $36 + b = 40$ and $b = 4$. Final answers: - Problem 1: $a=3$, $b=-4$ - Problem 2: $a=2$, $b=-7$ - Problem 3: $a=-\frac{37}{5}$, $b=-\frac{4}{5}$ - Problem 4: $a=3$, $b=4$