Simplify ((v^2-9)/(3v^2+12))÷((v+3)/(v^2+4))

$\frac{v^{2} - 9}{3 v^{2} + 12} \div \frac{v + 3}{v^{2} + 4}$

To divide by a fraction, multiply by its reciprocal.

$\frac{v^{2} - 9}{3 v^{2} + 12} \cdot \frac{v^{2} + 4}{v + 3}$

Simplify the numerator.

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

$\frac{v^{2} - 3^{2}}{3 v^{2} + 12} \cdot \frac{v^{2} + 4}{v + 3}$

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

$\frac{(v + 3) (v - 3)}{3 v^{2} + 12} \cdot \frac{v^{2} + 4}{v + 3}$

Simplify terms.

Factor $3$ out of $3 v^{2} + 12$ .

Factor $3$ out of $3 v^{2}$ .

$\frac{(v + 3) (v - 3)}{3 (v^{2}) + 12} \cdot \frac{v^{2} + 4}{v + 3}$

Factor $3$ out of $12$ .

$\frac{(v + 3) (v - 3)}{3 v^{2} + 3 \cdot 4} \cdot \frac{v^{2} + 4}{v + 3}$

Factor $3$ out of $3 v^{2} + 3 \cdot 4$ .

$\frac{(v + 3) (v - 3)}{3 (v^{2} + 4)} \cdot \frac{v^{2} + 4}{v + 3}$

Cancel the common factor of $v + 3$ .

Cancel the common factor.

$\frac{(v + 3) (v - 3)}{3 (v^{2} + 4)} \cdot \frac{v^{2} + 4}{v + 3}$

Rewrite the expression.

$\frac{v - 3}{3 (v^{2} + 4)} (v^{2} + 4)$

Cancel the common factor of $v^{2} + 4$ .

Factor $v^{2} + 4$ out of $3 (v^{2} + 4)$ .

$\frac{v - 3}{(v^{2} + 4) \cdot 3} (v^{2} + 4)$

Cancel the common factor.

$\frac{v - 3}{(v^{2} + 4) \cdot 3} (v^{2} + 4)$

Rewrite the expression.

$\frac{v - 3}{3}$