1. **State the problem:** Find the vertical asymptotes of the function $$f(x) = \frac{x^2 + 3}{(x^2 - 1)(x^2 - 25)}.$$\n\n2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points.\n\n3. **Set the denominator equal to zero:**\n$$ (x^2 - 1)(x^2 - 25) = 0 $$\n\n4. **Solve each factor:**\n$$ x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1 $$\n$$ x^2 - 25 = 0 \implies x^2 = 25 \implies x = \pm 5 $$\n\n5. **Check numerator at these points:**\n$$ x^2 + 3 $$ is always positive (since $x^2 \geq 0$), so numerator is not zero at $x = \pm 1, \pm 5$.\n\n6. **Conclusion:** The function has vertical asymptotes at $$x = -5, -1, 1, 5.$$\n\n7. **Order from leftmost to rightmost:** $$x = -5, -1, 1, 5.$$