1. **State the problem:** We have a polynomial $$f(x) = ax^4 + 8x^3 + 8b + 7$$ where $$a$$ and $$b$$ are constants. We know that $$f(x)$$ is divisible by $$2x + 5$$, and we want to find the remainder when $$f(x)$$ is divided by $$2x - 5$$. 2. **Use the Remainder Theorem:** If a polynomial $$f(x)$$ is divisible by a linear polynomial $$2x + 5$$, then $$f\left(-\frac{5}{2}\right) = 0$$. 3. **Apply the divisibility condition:** Substitute $$x = -\frac{5}{2}$$ into $$f(x)$$: $$f\left(-\frac{5}{2}\right) = a\left(-\frac{5}{2}\right)^4 + 8\left(-\frac{5}{2}\right)^3 + 8b + 7 = 0$$ Calculate powers: $$\left(-\frac{5}{2}\right)^4 = \frac{625}{16}$$ $$\left(-\frac{5}{2}\right)^3 = -\frac{125}{8}$$ So, $$f\left(-\frac{5}{2}\right) = a \cdot \frac{625}{16} + 8 \cdot \left(-\frac{125}{8}\right) + 8b + 7 = 0$$ Simplify: $$a \cdot \frac{625}{16} - 125 + 8b + 7 = 0$$ $$\Rightarrow \frac{625}{16}a + 8b - 118 = 0$$ 4. **Express $$8b$$ in terms of $$a$$:** $$8b = 118 - \frac{625}{16}a$$ 5. **Find the remainder when dividing by $$2x - 5$$:** By the Remainder Theorem, the remainder is $$f\left(\frac{5}{2}\right)$$. Calculate: $$f\left(\frac{5}{2}\right) = a\left(\frac{5}{2}\right)^4 + 8\left(\frac{5}{2}\right)^3 + 8b + 7$$ Calculate powers: $$\left(\frac{5}{2}\right)^4 = \frac{625}{16}$$ $$\left(\frac{5}{2}\right)^3 = \frac{125}{8}$$ So, $$f\left(\frac{5}{2}\right) = a \cdot \frac{625}{16} + 8 \cdot \frac{125}{8} + 8b + 7 = \frac{625}{16}a + 125 + 8b + 7$$ Simplify constants: $$= \frac{625}{16}a + 8b + 132$$ Substitute $$8b$$ from step 4: $$= \frac{625}{16}a + \left(118 - \frac{625}{16}a\right) + 132 = 118 + 132 = 250$$ 6. **Final answer:** The remainder when $$f(x)$$ is divided by $$2x - 5$$ is $$250$$. **Answer: C. 250**