Divide ((x^2-4)/(x-3))÷((x^2+7x-18)/(x+9))

$\frac{x^{2} - 4}{x - 3} \div \frac{x^{2} + 7 x - 18}{x + 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 4}{x - 3} \cdot \frac{x + 9}{x^{2} + 7 x - 18}$

Step 2

Simplify the numerator.

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

$\frac{x^{2} - 2^{2}}{x - 3} \cdot \frac{x + 9}{x^{2} + 7 x - 18}$

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 $b = 2$ .

$\frac{(x + 2) (x - 2)}{x - 3} \cdot \frac{x + 9}{x^{2} + 7 x - 18}$

Step 3

Factor $x^{2} + 7 x - 18$ 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 $- 18$ and whose sum is $7$ .

$- 2, 9$

Write the factored form using these integers.

$\frac{(x + 2) (x - 2)}{x - 3} \cdot \frac{x + 9}{(x - 2) (x + 9)}$

Step 4

Cancel the common factor of $x - 2$ .

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

$\frac{(x - 2) (x + 2)}{x - 3} \cdot \frac{x + 9}{(x - 2) (x + 9)}$

Cancel the common factor.

$\frac{(x - 2) (x + 2)}{x - 3} \cdot \frac{x + 9}{(x - 2) (x + 9)}$

Rewrite the expression.

$\frac{x + 2}{x - 3} \cdot \frac{x + 9}{x + 9}$

Step 5

Multiply $\frac{x + 2}{x - 3}$ by $\frac{x + 9}{x + 9}$ .

$\frac{(x + 2) (x + 9)}{(x - 3) (x + 9)}$

Step 6

Cancel the common factor of $x + 9$ .

Cancel the common factor.

$\frac{(x + 2) (x + 9)}{(x - 3) (x + 9)}$

Rewrite the expression.

$\frac{x + 2}{x - 3}$

Divide ((x^2-4)/(x-3))÷((x^2+7x-18)/(x+9))

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