Question: Divide and check the following.
$\frac{32a^3 + 8a^2 - 7a - 16}{-8a}$
$32a^3 + 8a^2 - 7a - 16 = \square$
(Simplify your answer. Type exponential notation with positive exponents. Use integers or fractions for any numbers in the expression.)
1. **State the problem:** We need to divide the polynomial $32a^3 + 8a^2 - 7a - 16$ by $-8a$ and simplify the result.
2. **Write the division as separate terms:**
$$\frac{32a^3}{-8a} + \frac{8a^2}{-8a} - \frac{7a}{-8a} - \frac{16}{-8a}$$
3. **Simplify each term individually:**
- For $\frac{32a^3}{-8a}$, divide coefficients and subtract exponents:
$$\frac{32}{-8} a^{3-1} = -4a^2$$
- For $\frac{8a^2}{-8a}$:
$$\frac{8}{-8} a^{2-1} = -1a = -a$$
- For $\frac{7a}{-8a}$:
$$\frac{7}{-8} a^{1-1} = -\frac{7}{8}$$
- For $\frac{16}{-8a}$:
$$\frac{16}{-8} \cdot \frac{1}{a} = -2 \cdot \frac{1}{a} = -\frac{2}{a}$$
4. **Combine all simplified terms:**
$$-4a^2 - a - \frac{7}{8} - \frac{2}{a}$$
5. **Final answer:**
$$\boxed{-4a^2 - a - \frac{7}{8} - \frac{2}{a}}$$
This is the simplified form of the division.