1. **State the problem:** Solve the equation $ (x+1)(x+3)(x+5)(x+7) = 9 $. 2. **Rewrite the equation:** Notice the terms are symmetric around the middle. Group them as $[(x+1)(x+7)] \times [(x+3)(x+5)] = 9$. 3. **Simplify each group:** $$ (x+1)(x+7) = x^2 + 8x + 7 $$ $$ (x+3)(x+5) = x^2 + 8x + 15 $$ 4. **Substitute:** Let $y = x^2 + 8x$, then the equation becomes: $$ (y + 7)(y + 15) = 9 $$ 5. **Expand:** $$ y^2 + 15y + 7y + 105 = 9 $$ $$ y^2 + 22y + 105 = 9 $$ 6. **Bring all terms to one side:** $$ y^2 + 22y + 105 - 9 = 0 $$ $$ y^2 + 22y + 96 = 0 $$ 7. **Solve quadratic in $y$:** Use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=22$, $c=96$. Calculate discriminant: $$ \Delta = 22^2 - 4 \times 1 \times 96 = 484 - 384 = 100 $$ So, $$ y = \frac{-22 \pm \sqrt{100}}{2} = \frac{-22 \pm 10}{2} $$ Two solutions for $y$: $$ y_1 = \frac{-22 + 10}{2} = \frac{-12}{2} = -6 $$ $$ y_2 = \frac{-22 - 10}{2} = \frac{-32}{2} = -16 $$ 8. **Recall $y = x^2 + 8x$, solve for $x$:** For $y = -6$: $$ x^2 + 8x = -6 $$ $$ x^2 + 8x + 6 = 0 $$ For $y = -16$: $$ x^2 + 8x = -16 $$ $$ x^2 + 8x + 16 = 0 $$ 9. **Solve each quadratic:** For $x^2 + 8x + 6 = 0$: Discriminant: $$ \Delta = 8^2 - 4 \times 1 \times 6 = 64 - 24 = 40 $$ Solutions: $$ x = \frac{-8 \pm \sqrt{40}}{2} = \frac{-8 \pm 2\sqrt{10}}{2} = -4 \pm \sqrt{10} $$ For $x^2 + 8x + 16 = 0$: Discriminant: $$ \Delta = 8^2 - 4 \times 1 \times 16 = 64 - 64 = 0 $$ One solution: $$ x = \frac{-8}{2} = -4 $$ 10. **Final solutions:** $$ x = -4 + \sqrt{10}, \quad x = -4 - \sqrt{10}, \quad x = -4 $$