1. **Problem Statement:** Given the functions $f(x) = x^3 - x^2 + ax + b$ and $g(x) = x^2 - 2x - 8$, if $g(x)$ is a factor of $f(x)$, prove that $2a - 3b = 4$. 2. **Understanding the problem:** To say $g(x)$ is a factor of $f(x)$ means that when $f(x)$ is divided by $g(x)$, the remainder is zero. Since $g(x)$ is quadratic, the remainder would be a linear polynomial or constant. For $g(x)$ to be a factor, $f(x)$ must be zero at the roots of $g(x)$. 3. **Find roots of $g(x)$:** Solve $g(x) = 0$: $$x^2 - 2x - 8 = 0$$ Using the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2}$$ So the roots are: $$x = 4 \quad \text{and} \quad x = -2$$ 4. **Apply factor condition:** Since $g(x)$ divides $f(x)$, $f(4) = 0$ and $f(-2) = 0$. Calculate $f(4)$: $$f(4) = 4^3 - 4^2 + a \cdot 4 + b = 64 - 16 + 4a + b = 48 + 4a + b$$ Set equal to zero: $$48 + 4a + b = 0 \implies 4a + b = -48 \quad (1)$$ Calculate $f(-2)$: $$f(-2) = (-2)^3 - (-2)^2 + a(-2) + b = -8 - 4 - 2a + b = -12 - 2a + b$$ Set equal to zero: $$-12 - 2a + b = 0 \implies -2a + b = 12 \quad (2)$$ 5. **Solve the system of equations:** From (1) and (2): $$4a + b = -48$$ $$-2a + b = 12$$ Subtract (2) from (1): $$(4a + b) - (-2a + b) = -48 - 12$$ $$4a + b + 2a - b = -60$$ $$6a = -60 \implies a = -10$$ Substitute $a = -10$ into (1): $$4(-10) + b = -48 \implies -40 + b = -48 \implies b = -8$$ 6. **Verify the required expression:** Calculate $2a - 3b$: $$2(-10) - 3(-8) = -20 + 24 = 4$$ **Hence proved:** $$2a - 3b = 4$$