1. **State the problem:** We want to find all integers $n$ such that $$(n + 3)(n - 7) \leq 0.$$ 2. **Understand the inequality:** The product of two factors is less than or equal to zero. This means the product is either zero or negative. 3. **Find the roots:** Set each factor equal to zero: $$n + 3 = 0 \implies n = -3$$ $$n - 7 = 0 \implies n = 7$$ 4. **Determine intervals:** The critical points split the number line into three intervals: - $(-\infty, -3)$ - $[-3, 7]$ - $(7, \infty)$ 5. **Test each interval:** - For $n < -3$, pick $n = -4$: $( -4 + 3)( -4 - 7) = (-1)(-11) = 11 > 0$ (positive) - For $-3 \leq n \leq 7$, pick $n = 0$: $(0 + 3)(0 - 7) = 3 \times (-7) = -21 < 0$ (negative) - For $n > 7$, pick $n = 8$: $(8 + 3)(8 - 7) = 11 \times 1 = 11 > 0$ (positive) 6. **Include points where product is zero:** At $n = -3$ or $n = 7$, the product is zero, so these points satisfy the inequality. 7. **Conclusion:** The inequality holds for all integers $n$ in the interval $$[-3, 7].$$ 8. **Count the integers:** The integers from $-3$ to $7$ inclusive are: $$-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7$$ There are $7 - (-3) + 1 = 11$ integers. **Final answer:** $$\boxed{11}$$