Divide ((x-4)/(x+3))÷((x^2-16)/(x^2+6x+9))

$\frac{x - 4}{x + 3} \div \frac{x^{2} - 16}{x^{2} + 6 x + 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x - 4}{x + 3} \cdot \frac{x^{2} + 6 x + 9}{x^{2} - 16}$

Step 2

Factor using the perfect square rule.

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

$\frac{x - 4}{x + 3} \cdot \frac{x^{2} + 6 x + 3^{2}}{x^{2} - 16}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$\frac{x - 4}{x + 3} \cdot \frac{x^{2} + 2 \cdot x \cdot 3 + 3^{2}}{x^{2} - 16}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

$\frac{x - 4}{x + 3} \cdot \frac{{(x + 3)}^{2}}{x^{2} - 16}$

Step 3

Simplify the denominator.

Rewrite $16$ as $4^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

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

Step 4

Cancel the common factor of $x - 4$ .

Factor $x - 4$ out of $(x + 4) (x - 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

Cancel the common factor of $x + 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 3}{x + 4}$