Divide ((x^3)/(x^2-4))÷((2x)/(x^2-7x+10))

$\frac{x^{3}}{x^{2} - 4} \div \frac{2 x}{x^{2} - 7 x + 10}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{3}}{x^{2} - 4} \cdot \frac{x^{2} - 7 x + 10}{2 x}$

Cancel the common factor of $x$ .

Factor $x$ out of $x^{3}$ .

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

Factor $x$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $\frac{x^{2}}{x^{2} - 4}$ by $\frac{x^{2} - 7 x + 10}{2}$ .

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

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 $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

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

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

Simplify the denominator.

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

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

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 $b = 2$ .

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Move $2$ to the left of $x + 2$ .

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

Divide ((x^3)/(x^2-4))÷((2x)/(x^2-7x+10))

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