1. **State the problem:** We are given the function $$h(x) = \frac{6x^2 - 23x + 20}{2x - 5}$$ and need to express it in the form $$ax + b$$ where $$a, b \in \mathbb{Z}$$. 2. **Formula and approach:** To express $$h(x)$$ as a linear function, we perform polynomial division of the numerator by the denominator. 3. **Perform polynomial division:** Divide $$6x^2 - 23x + 20$$ by $$2x - 5$$. - First, divide the leading terms: $$\frac{6x^2}{2x} = 3x$$. - Multiply $$3x$$ by $$2x - 5$$: $$3x \times (2x - 5) = 6x^2 - 15x$$. - Subtract this from the numerator: $$\begin{aligned} & (6x^2 - 23x + 20) - (6x^2 - 15x) \\ &= 6x^2 - 23x + 20 - 6x^2 + 15x \\ &= -8x + 20 \end{aligned}$$ - Next, divide the new leading term by the divisor's leading term: $$\frac{-8x}{2x} = -4$$. - Multiply $$-4$$ by $$2x - 5$$: $$-4 \times (2x - 5) = -8x + 20$$. - Subtract this from the remainder: $$(-8x + 20) - (-8x + 20) = 0$$. 4. **Result:** The quotient is $$3x - 4$$ with remainder 0, so $$h(x) = 3x - 4$$. 5. **Explanation:** Polynomial division shows that the rational function simplifies exactly to the linear function $$3x - 4$$. **Final answer:** $$h(x) = 3x - 4$$