1. **State the problem:** Find the zeros of the function $$f(x) = \frac{x^2 + 14x + 49}{x^2 + 1}$$. 2. **Recall the rule for zeros of a rational function:** The zeros occur where the numerator is zero and the denominator is not zero. 3. **Set the numerator equal to zero:** $$x^2 + 14x + 49 = 0$$ 4. **Factor the numerator:** $$x^2 + 14x + 49 = (x + 7)^2$$ 5. **Solve for x:** $$ (x + 7)^2 = 0 \implies x + 7 = 0 \implies x = -7 $$ 6. **Check the denominator at this x-value:** $$x^2 + 1 = (-7)^2 + 1 = 49 + 1 = 50 \neq 0$$ 7. **Conclusion:** The function has a zero at $$x = -7$$. **Final answer:** $$x = -7$$