Factor 9x^2-1

$9 x^{2} - 1$

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

${(3 x)}^{2} - 1$

Rewrite $1$ as $1^{2}$ .

${(3 x)}^{2} - 1^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

$(3 x + 1) (3 x - 1)$

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