Factor the expression below. $$ \begin{array}{l}x^{2}-81 \ \text { A. }(x+27)(x-3) \ \text { B. }(x-27)(x-3) \ \text { C. }(x-9)(x-9) \ \text { D. }(x+9)(x-9)\end{array} $$

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The expression $$ x^2 - 81 $$ factors to $$ (x + 9)(x - 9) $$. The correct answer is D. The expression given is $$ x^2 - 81 $$. This expression is a difference of squares, which can be factored using the formula: $$ a^2 - b^2 = (a - b)(a + b) $$ In this case, we can identify $$ a = x $$ and $$ b = 9 $$ since $$ 81 = 9^2 $$. Therefore, we can apply the difference of squares formula: $$ x^2 - 81 = (x - 9)(x + 9) $$ Now, let's compare this with the provided options: - A. $$ (x + 27)(x - 3) $$ - B. $$ (x - 27)(x - 3) $$ - C. $$ (x - 9)(x - 9) $$ - D. $$ (x + 9)(x - 9) $$ The correct factorization is: $$ (x + 9)(x - 9) $$ Thus, the answer is: $$ \boxed{D} $$

Factor the expression below. \[ \begin{array}{l}x^{2}-81 \\ \text { A. }(x+27)(x-3) \\ \text { B. }(x-27)(x-3) \\ \text { C. }(x-9)(x-9) \\ \text { D. }(x+9)(x-9)\end{array} \]

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Factor the expression below. \[ \begin{array}{l}x^{2}-81 \\ \text { A. }(x+27)(x-3) \\ \text { B. }(x-27)(x-3) \\ \text { C. }(x-9)(x-9) \\ \text { D. }(x+9)(x-9)\end{array} \]

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