Divide ((x+8)/(x^2-64))÷((x-4)/(x^2+x-72))

$\frac{x + 8}{x^{2} - 64} \div \frac{x - 4}{x^{2} + x - 72}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x + 8}{x^{2} - 64} \cdot \frac{x^{2} + x - 72}{x - 4}$

Simplify the denominator.

Rewrite $64$ as $8^{2}$ .

$\frac{x + 8}{x^{2} - 8^{2}} \cdot \frac{x^{2} + x - 72}{x - 4}$

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 $b = 8$ .

$\frac{x + 8}{(x + 8) (x - 8)} \cdot \frac{x^{2} + x - 72}{x - 4}$

Cancel the common factor of $x + 8$ .

Cancel the common factor.

$\frac{x + 8}{(x + 8) (x - 8)} \cdot \frac{x^{2} + x - 72}{x - 4}$

Rewrite the expression.

$\frac{1}{x - 8} \cdot \frac{x^{2} + x - 72}{x - 4}$

Factor $x^{2} + x - 72$ 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 $- 72$ and whose sum is $1$ .

$- 8, 9$

Write the factored form using these integers.

$\frac{1}{x - 8} \cdot \frac{(x - 8) (x + 9)}{x - 4}$

Cancel the common factor of $x - 8$ .

Cancel the common factor.

$\frac{1}{x - 8} \cdot \frac{(x - 8) (x + 9)}{x - 4}$

Rewrite the expression.

$\frac{x + 9}{x - 4}$

Divide ((x+8)/(x^2-64))÷((x-4)/(x^2+x-72))

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