1. **Stating the problem:** We have a polynomial $f(x)$ such that $$f(x) = (x+1)(x-2)h(x) + ax + b,$$ where $a$ and $b$ are real constants. It leaves a remainder of 2 when divided by $(x+1)$ and a remainder of 23 when divided by $(x-2)$. We need to find $a$ and $b$. Also, $f(x)$ is a cubic polynomial with leading coefficient 3 and $f(1) = 8$. We want to find an expression for $f(x)$. --- 2. **Using the Remainder Theorem:** - When dividing by $(x+1)$, the remainder is $f(-1) = 2$. - When dividing by $(x-2)$, the remainder is $f(2) = 23$. Since $$f(x) = (x+1)(x-2)h(x) + ax + b,$$ plugging in $x = -1$ and $x = 2$ gives: $$f(-1) = a(-1) + b = -a + b = 2,$$ $$f(2) = 2a + b = 23.$$ --- 3. **Solving the system for $a$ and $b$:** From the two equations: $$-a + b = 2 \quad (1)$$ $$2a + b = 23 \quad (2)$$ Subtract (1) from (2): $$(2a + b) - (-a + b) = 23 - 2$$ $$2a + b + a - b = 21$$ $$3a = 21$$ $$a = 7 \quad \text{(M1)}$$ Substitute $a=7$ into (1): $$-7 + b = 2$$ $$b = 9 \quad \text{(A1)}$$ --- 4. **Finding the cubic polynomial $f(x)$:** Given $f(x)$ is cubic with leading coefficient 3, write: $$f(x) = 3x^3 + cx^2 + dx + e,$$ where $c, d, e$ are constants to find. We know from the problem: $$f(x) = (x+1)(x-2)h(x) + ax + b,$$ and since $a=7$, $b=9$, the remainder part is $7x + 9$. Because $(x+1)(x-2)$ is quadratic, $h(x)$ must be linear for $f(x)$ to be cubic: Let $$h(x) = px + q,$$ then $$f(x) = (x+1)(x-2)(px + q) + 7x + 9.$$ Expand $(x+1)(x-2) = x^2 - x - 2$: $$f(x) = (x^2 - x - 2)(px + q) + 7x + 9.$$ Multiply out: $$f(x) = p x^3 + q x^2 - p x^2 - q x - 2 p x - 2 q + 7 x + 9.$$ Group like terms: $$f(x) = p x^3 + (q - p) x^2 + (-q - 2p + 7) x + (-2 q + 9).$$ Since leading coefficient is 3: $$p = 3 \quad \text{(M1)}$$ So, $$f(x) = 3 x^3 + (q - 3) x^2 + (-q - 6 + 7) x + (-2 q + 9)$$ $$= 3 x^3 + (q - 3) x^2 + (-q + 1) x + (-2 q + 9).$$ --- 5. **Use $f(1) = 8$ to find $q$:** $$f(1) = 3(1)^3 + (q - 3)(1)^2 + (-q + 1)(1) + (-2 q + 9) = 8.$$ Simplify: $$3 + q - 3 - q + 1 - 2 q + 9 = 8,$$ Combine like terms: $$3 - 3 + 1 + 9 + q - q - 2 q = 8,$$ $$10 - 2 q = 8,$$ Solve for $q$: $$-2 q = 8 - 10 = -2,$$ $$q = 1 \quad \text{(A1)}$$ --- 6. **Final expression for $f(x)$:** Substitute $q=1$: $$f(x) = 3 x^3 + (1 - 3) x^2 + (-1 + 1) x + (-2(1) + 9),$$ Simplify: $$f(x) = 3 x^3 - 2 x^2 + 0 x + 7,$$ or $$\boxed{f(x) = 3 x^3 - 2 x^2 + 7}.$$