Simplify ((x-3)/(x^2+4x-21))÷((x^2-25)/(x-5))

$\frac{x - 3}{x^{2} + 4 x - 21} \div \frac{x^{2} - 25}{x - 5}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x - 3}{x^{2} + 4 x - 21} \cdot \frac{x - 5}{x^{2} - 25}$

Step 2

Factor $x^{2} + 4 x - 21$ 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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

$\frac{x - 3}{(x - 3) (x + 7)} \cdot \frac{x - 5}{x^{2} - 25}$

Step 3

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{x - 3}{(x - 3) (x + 7)} \cdot \frac{x - 5}{x^{2} - 25}$

Rewrite the expression.

$\frac{1}{x + 7} \cdot \frac{x - 5}{x^{2} - 25}$

Step 4

Simplify the denominator.

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

$\frac{1}{x + 7} \cdot \frac{x - 5}{x^{2} - 5^{2}}$

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 $b = 5$ .

$\frac{1}{x + 7} \cdot \frac{x - 5}{(x + 5) (x - 5)}$

Step 5

Simplify terms.

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$\frac{1}{x + 7} \cdot \frac{x - 5}{(x + 5) (x - 5)}$

Rewrite the expression.

$\frac{1}{x + 7} \cdot \frac{1}{x + 5}$

Multiply $\frac{1}{x + 7}$ by $\frac{1}{x + 5}$ .

$\frac{1}{(x + 7) (x + 5)}$

Simplify ((x-3)/(x^2+4x-21))÷((x^2-25)/(x-5))

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