It appears there might be a misunderstanding in the problem statement as presented. If you meant to solve $x^2 = 81$ (which is a quadratic equation), here's how you solve it by factoring:

1. Rewrite the equation in standard quadratic form: $x^2 - 81 = 0$

2. Recognize that this is a difference of squares, which can be factored as: $(x - 9)(x + 9) = 0$

3. Set each factor equal to zero and solve for $x$: $x - 9 = 0 \quad \text{or} \quad x + 9 = 0$

   $x = 9 \quad \text{or} \quad x = -9$

Thus, the solutions are $x = 9$ and $x = -9$.

Would you like any further details or have any questions about this solution?

Here are 8 related questions to expand your understanding:

1. How would you solve the equation $x^2 - 25 = 0$ by factoring?
2. What are the solutions to the equation $x^2 - 16 = 0$?
3. How do you factor and solve $x^2 - 49 = 0$?
4. What are the roots of the equation $x^2 - 64 = 0$?
5. How would you solve $x^2 - 4 = 0$ by factoring?
6. What are the solutions to the equation $x^2 - 1 = 0$?
7. How do you factor and solve $x^2 - 100 = 0$?
8. What are the roots of the equation $x^2 - 121 = 0$?

Tip: When dealing with quadratic equations, always look for patterns such as the difference of squares, perfect square trinomials, and common factors to simplify the factoring process.