Factor 9y^2-1

$9 y^{2} - 1$

Step 1

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

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

Step 2

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

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

Step 3

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

$(3 y + 1) (3 y - 1)$