Cubic Polynomial Quotient

1. **Problem statement:** We have a cubic polynomial $f(x)$ such that: - When divided by $x+1$, remainder is $-3$. - When divided by $x-2$, remainder is $66$. - $f(x)$ is divisible by $2x^2 + 5x + 4$. We need to find the quotient when $f(x)$ is divided by $2x^2 + 5x + 4$. 2. **Express $f(x)$:** Since $f(x)$ is cubic and divisible by $2x^2 + 5x + 4$, the quotient must be linear: $$f(x) = (2x^2 + 5x + 4)(Ax + B)$$ where $A$ and $B$ are constants to be found. 3. **Expand $f(x)$:** $$f(x) = (2x^2 + 5x + 4)(Ax + B) = 2A x^3 + 5A x^2 + 4A x + 2B x^2 + 5B x + 4B$$ Combine like terms: $$f(x) = 2A x^3 + (5A + 2B) x^2 + (4A + 5B) x + 4B$$ 4. **Use remainder conditions:** - Remainder when divided by $x+1$ is $f(-1) = -3$: $$f(-1) = 2A(-1)^3 + (5A + 2B)(-1)^2 + (4A + 5B)(-1) + 4B = -2A + 5A + 2B - 4A - 5B + 4B$$ Simplify: $$-2A + 5A - 4A + 2B - 5B + 4B = (-2A + 5A - 4A) + (2B - 5B + 4B) = (-1A) + (1B) = -A + B$$ Set equal to $-3$: $$-A + B = -3$$ - Remainder when divided by $x-2$ is $f(2) = 66$: $$f(2) = 2A(2)^3 + (5A + 2B)(2)^2 + (4A + 5B)(2) + 4B = 2A imes 8 + (5A + 2B) imes 4 + (4A + 5B) imes 2 + 4B$$ Calculate: $$16A + 20A + 8B + 8A + 10B + 4B = (16A + 20A + 8A) + (8B + 10B + 4B) = 44A + 22B$$ Set equal to $66$: $$44A + 22B = 66$$ 5. **Solve the system:** From first equation: $$B = A - 3$$ Substitute into second: $$44A + 22(A - 3) = 66$$ $$44A + 22A - 66 = 66$$ $$66A = 132$$ $$A = 2$$ Then: $$B = 2 - 3 = -1$$ 6. **Write the quotient:** $$\boxed{2x - 1}$$ **Final answer:** The quotient when $f(x)$ is divided by $2x^2 + 5x + 4$ is $2x - 1$.