Simplify the Radical Expression ((x^2-25)/(x^2+7x+10))÷((x-5)/(4(x+2)))

$\frac{x^{2} - 25}{x^{2} + 7 x + 10} \div \frac{x - 5}{4 (x + 2)}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 25}{x^{2} + 7 x + 10} \cdot \frac{4 (x + 2)}{x - 5}$

Simplify the numerator.

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

$\frac{x^{2} - 5^{2}}{x^{2} + 7 x + 10} \cdot \frac{4 (x + 2)}{x - 5}$

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 $b = 5$ .

$\frac{(x + 5) (x - 5)}{x^{2} + 7 x + 10} \cdot \frac{4 (x + 2)}{x - 5}$

Factor $x^{2} + 7 x + 10$ 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 $10$ and whose sum is $7$ .

$2, 5$

Write the factored form using these integers.

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

Cancel the common factor of $x - 5$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 2$ .

Factor $x + 2$ out of $4 (x + 2)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 5}{x + 5} \cdot 4$

Combine $\frac{x + 5}{x + 5}$ and $4$ .

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

Cancel the common factor of $x + 5$ .

Cancel the common factor.

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

Divide $4$ by $1$ .

$4$

Simplify the Radical Expression ((x^2-25)/(x^2+7x+10))÷((x-5)/(4(x+2)))

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