1. **Problem statement:** Find the cubic polynomial $p(x)$ given that when divided by $x$, the remainder is 1; when divided by $x-2$, the remainder is 9; when divided by $x+2$, the remainder is -15; and $p(x)$ is divisible by $x-1$. 2. **Recall the Remainder Theorem:** If a polynomial $p(x)$ is divided by $x-a$, the remainder is $p(a)$. 3. **Apply the Remainder Theorem:** - Remainder when divided by $x$ means $p(0) = 1$. - Remainder when divided by $x-2$ means $p(2) = 9$. - Remainder when divided by $x+2$ means $p(-2) = -15$. - Since $p(x)$ is divisible by $x-1$, $p(1) = 0$. 4. **Assume general form:** Let $p(x) = ax^3 + bx^2 + cx + d$. 5. **Use the conditions:** - $p(0) = d = 1$ - $p(1) = a + b + c + d = 0$ - $p(2) = 8a + 4b + 2c + d = 9$ - $p(-2) = -8a + 4b - 2c + d = -15$ 6. **Substitute $d=1$ into equations:** - $a + b + c + 1 = 0 \\ \Rightarrow a + b + c = -1$ - $8a + 4b + 2c + 1 = 9 \\ \Rightarrow 8a + 4b + 2c = 8$ - $-8a + 4b - 2c + 1 = -15 \\ \Rightarrow -8a + 4b - 2c = -16$ 7. **Solve the system:** From equations: $$\begin{cases} a + b + c = -1 \\ 8a + 4b + 2c = 8 \\ -8a + 4b - 2c = -16 \end{cases}$$ Add the last two equations: $$ (8a - 8a) + (4b + 4b) + (2c - 2c) = 8 - 16 \\ 0 + 8b + 0 = -8 \\ 8b = -8 \\ b = -1$$ Substitute $b = -1$ into first equation: $$ a -1 + c = -1 \\ a + c = 0 \\ c = -a$$ Substitute $b = -1$ and $c = -a$ into second equation: $$ 8a + 4(-1) + 2(-a) = 8 \\ 8a - 4 - 2a = 8 \\ 6a - 4 = 8 \\ 6a = 12 \\ a = 2$$ Then $c = -a = -2$. 8. **Write the polynomial:** $$ p(x) = 2x^3 - x^2 - 2x + 1 $$ 9. **Check divisibility by $x-1$:** $$ p(1) = 2 - 1 - 2 + 1 = 0 $$ Correct. 10. **Answer to part (b):** Claim: All roots of $p(x) = 0$ are integers. Check rational roots using Rational Root Theorem: possible roots are factors of 1 over factors of 2, i.e., $\pm1, \pm\frac{1}{2}$. Test $x=1$: root. Divide $p(x)$ by $(x-1)$: $$ p(x) = (x-1)(2x^2 + x - 1) $$ Factor quadratic: $$ 2x^2 + x - 1 = (2x - 1)(x + 1) $$ Roots are $x=1$, $x=\frac{1}{2}$, and $x=-1$. Since $\frac{1}{2}$ is not an integer, the claim is false. **Final answers:** (a) $p(x) = 2x^3 - x^2 - 2x + 1$ (b) No, not all roots are integers because one root is $\frac{1}{2}$, which is rational but not integer.