Factor (x+3)^2-9

${(x + 3)}^{2} - 9$

Step 1

Rewrite $9$ as $3^{2}$ .

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

Step 2

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

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

Step 3

Simplify.

Add $3$ and $3$ .

$(x + 6) (x + 3 - 3)$

Subtract $3$ from $3$ .

$(x + 6) (x + 0)$

Add $x$ and $0$ .

$(x + 6) x$

Factor (x+3)^2-9

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