Simplify ((2m^2+3m)/(2m^2-7m-15))÷((8m^2+12m)/(m^2-25))

$\frac{2 m^{2} + 3 m}{2 m^{2} - 7 m - 15} \div \frac{8 m^{2} + 12 m}{m^{2} - 25}$

To divide by a fraction, multiply by its reciprocal.

$\frac{2 m^{2} + 3 m}{2 m^{2} - 7 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor $m$ out of $2 m^{2} + 3 m$ .

Factor $m$ out of $2 m^{2}$ .

$\frac{m (2 m) + 3 m}{2 m^{2} - 7 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor $m$ out of $3 m$ .

$\frac{m (2 m) + m \cdot 3}{2 m^{2} - 7 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor $m$ out of $m (2 m) + m \cdot 3$ .

$\frac{m (2 m + 3)}{2 m^{2} - 7 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 m$ .

$\frac{m (2 m + 3)}{2 m^{2} - 7 (m) - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Rewrite $- 7$ as $3$ plus $- 10$

$\frac{m (2 m + 3)}{2 m^{2} + (3 - 10) m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Apply the distributive property.

$\frac{m (2 m + 3)}{2 m^{2} + 3 m - 10 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{m (2 m + 3)}{(2 m^{2} + 3 m) - 10 m - 15} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor out the greatest common factor (GCF) from each group.

$\frac{m (2 m + 3)}{m (2 m + 3) - 5 (2 m + 3)} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Factor the polynomial by factoring out the greatest common factor, $2 m + 3$ .

$\frac{m (2 m + 3)}{(2 m + 3) (m - 5)} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Cancel the common factor of $2 m + 3$ .

Cancel the common factor.

$\frac{m (2 m + 3)}{(2 m + 3) (m - 5)} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Rewrite the expression.

$\frac{m}{m - 5} \cdot \frac{m^{2} - 25}{8 m^{2} + 12 m}$

Simplify the numerator.

Rewrite $25$ as $5^{2}$ .

$\frac{m}{m - 5} \cdot \frac{m^{2} - 5^{2}}{8 m^{2} + 12 m}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = m$ and $b = 5$ .

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{8 m^{2} + 12 m}$

Simplify terms.

Factor $4 m$ out of $8 m^{2} + 12 m$ .

Factor $4 m$ out of $8 m^{2}$ .

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{4 m (2 m) + 12 m}$

Factor $4 m$ out of $12 m$ .

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{4 m (2 m) + 4 m (3)}$

Factor $4 m$ out of $4 m (2 m) + 4 m (3)$ .

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{4 m (2 m + 3)}$

Cancel the common factor of $m$ .

Factor $m$ out of $4 m (2 m + 3)$ .

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{m (4 (2 m + 3))}$

Cancel the common factor.

$\frac{m}{m - 5} \cdot \frac{(m + 5) (m - 5)}{m (4 (2 m + 3))}$

Rewrite the expression.

$\frac{1}{m - 5} \cdot \frac{(m + 5) (m - 5)}{4 (2 m + 3)}$

Cancel the common factor of $m - 5$ .

Factor $m - 5$ out of $(m + 5) (m - 5)$ .

$\frac{1}{m - 5} \cdot \frac{(m - 5) (m + 5)}{4 (2 m + 3)}$

Cancel the common factor.

$\frac{1}{m - 5} \cdot \frac{(m - 5) (m + 5)}{4 (2 m + 3)}$

Rewrite the expression.

$\frac{m + 5}{4 (2 m + 3)}$

Simplify ((2m^2+3m)/(2m^2-7m-15))÷((8m^2+12m)/(m^2-25))

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