Simplify ((x+3)/(4x^2-9))÷((x^2+7x+12)/(2x^2+7x-15))

$\frac{x + 3}{4 x^{2} - 9} \div \frac{x^{2} + 7 x + 12}{2 x^{2} + 7 x - 15}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x + 3}{4 x^{2} - 9} \cdot \frac{2 x^{2} + 7 x - 15}{x^{2} + 7 x + 12}$

Simplify the denominator.

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{x + 3}{{(2 x)}^{2} - 9} \cdot \frac{2 x^{2} + 7 x - 15}{x^{2} + 7 x + 12}$

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

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

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

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

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 $7 x$ .

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

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

$\frac{x + 3}{(2 x + 3) (2 x - 3)} \cdot \frac{2 x^{2} + (- 3 + 10) x - 15}{x^{2} + 7 x + 12}$

Apply the distributive property.

$\frac{x + 3}{(2 x + 3) (2 x - 3)} \cdot \frac{2 x^{2} - 3 x + 10 x - 15}{x^{2} + 7 x + 12}$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{x + 3}{(2 x + 3) (2 x - 3)} \cdot \frac{(2 x^{2} - 3 x) + 10 x - 15}{x^{2} + 7 x + 12}$

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

$\frac{x + 3}{(2 x + 3) (2 x - 3)} \cdot \frac{x (2 x - 3) + 5 (2 x - 3)}{x^{2} + 7 x + 12}$

Factor the polynomial by factoring out the greatest common factor, $2 x - 3$ .

$\frac{x + 3}{(2 x + 3) (2 x - 3)} \cdot \frac{(2 x - 3) (x + 5)}{x^{2} + 7 x + 12}$

Factor $x^{2} + 7 x + 12$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $12$ and whose sum is $7$ .

$3, 4$

Write the factored form using these integers.

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

Simplify terms.

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2 x - 3$ .

Factor $2 x - 3$ out of $(2 x + 3) (2 x - 3)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{2 x + 3} \cdot \frac{x + 5}{x + 4}$

Multiply $\frac{1}{2 x + 3}$ by $\frac{x + 5}{x + 4}$ .

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

Simplify ((x+3)/(4x^2-9))÷((x^2+7x+12)/(2x^2+7x-15))

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