1. **State the problem:** Find the equations of the vertical asymptotes of the function $$f(x) = \frac{x^2 + 2}{(x^2 - 9)(x^2 - 25)}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. **Factor the denominator:** $$x^2 - 9 = (x - 3)(x + 3)$$ $$x^2 - 25 = (x - 5)(x + 5)$$ So the denominator is: $$(x - 3)(x + 3)(x - 5)(x + 5)$$ 4. **Set denominator equal to zero to find potential vertical asymptotes:** $$(x - 3)(x + 3)(x - 5)(x + 5) = 0$$ This gives: $$x = 3, -3, 5, -5$$ 5. **Check numerator at these points:** $$x^2 + 2$$ is always positive (since $x^2 \geq 0$ and $+2$), so numerator is not zero at these points. 6. **Conclusion:** The vertical asymptotes are at $$x = -5, -3, 3, 5$$.