Divide ((x^2-25)/(x^2-10x+25))÷((x^2+10x+25)/(2x-10))

$\frac{x^{2} - 25}{x^{2} - 10 x + 25} \div \frac{x^{2} + 10 x + 25}{2 x - 10}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 25}{x^{2} - 10 x + 25} \cdot \frac{2 x - 10}{x^{2} + 10 x + 25}$

Step 2

Simplify the numerator.

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

$\frac{x^{2} - 5^{2}}{x^{2} - 10 x + 25} \cdot \frac{2 x - 10}{x^{2} + 10 x + 25}$

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 $b = 5$ .

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

Step 3

Factor using the perfect square rule.

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

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

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

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

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

Step 4

Cancel the common factor of $x - 5$ and ${(x - 5)}^{2}$ .

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

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

Cancel the common factors.

Factor $x - 5$ out of ${(x - 5)}^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 5}{x - 5} \cdot \frac{2 x - 10}{x^{2} + 10 x + 25}$

Step 5

Factor $2$ out of $2 x - 10$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 10$ .

$\frac{x + 5}{x - 5} \cdot \frac{2 x + 2 \cdot - 5}{x^{2} + 10 x + 25}$

Factor $2$ out of $2 x + 2 \cdot - 5$ .

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

Step 6

Factor using the perfect square rule.

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

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

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

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 5$ .

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

Step 7

Cancel the common factor of $x + 5$ .

Factor $x + 5$ out of ${(x + 5)}^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{x - 5} \cdot \frac{2 (x - 5)}{x + 5}$

Step 8

Cancel the common factor of $x - 5$ .

Factor $x - 5$ out of $2 (x - 5)$ .

$\frac{1}{x - 5} \cdot \frac{(x - 5) \cdot 2}{x + 5}$

Cancel the common factor.

$\frac{1}{x - 5} \cdot \frac{(x - 5) \cdot 2}{x + 5}$

Rewrite the expression.

$\frac{2}{x + 5}$

Divide ((x^2-25)/(x^2-10x+25))÷((x^2+10x+25)/(2x-10))

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