Divide ((x^2)/(x^2-x-42))÷((5x)/(x^2-36))

$\frac{x^{2}}{x^{2} - x - 42} \div \frac{5 x}{x^{2} - 36}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2}}{x^{2} - x - 42} \cdot \frac{x^{2} - 36}{5 x}$

Cancel the common factor of $x$ .

Factor $x$ out of $x^{2}$ .

$\frac{x \cdot x}{x^{2} - x - 42} \cdot \frac{x^{2} - 36}{5 x}$

Factor $x$ out of $5 x$ .

$\frac{x \cdot x}{x^{2} - x - 42} \cdot \frac{x^{2} - 36}{x \cdot 5}$

Cancel the common factor.

$\frac{x \cdot x}{x^{2} - x - 42} \cdot \frac{x^{2} - 36}{x \cdot 5}$

Rewrite the expression.

$\frac{x}{x^{2} - x - 42} \cdot \frac{x^{2} - 36}{5}$

Multiply $\frac{x}{x^{2} - x - 42}$ by $\frac{x^{2} - 36}{5}$ .

$\frac{x (x^{2} - 36)}{(x^{2} - x - 42) \cdot 5}$

Simplify the numerator.

Rewrite $36$ as $6^{2}$ .

$\frac{x (x^{2} - 6^{2})}{(x^{2} - x - 42) \cdot 5}$

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 $b = 6$ .

$\frac{x (x + 6) (x - 6)}{(x^{2} - x - 42) \cdot 5}$

Factor $x^{2} - x - 42$ 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 $- 42$ and whose sum is $- 1$ .

$- 7, 6$

Write the factored form using these integers.

$\frac{x (x + 6) (x - 6)}{(x - 7) (x + 6) \cdot 5}$

Cancel the common factor of $x + 6$ .

Cancel the common factor.

$\frac{x (x + 6) (x - 6)}{(x - 7) (x + 6) \cdot 5}$

Rewrite the expression.

$\frac{x (x - 6)}{(x - 7) \cdot 5}$

Move $5$ to the left of $x - 7$ .

$\frac{x (x - 6)}{5 (x - 7)}$

Divide ((x^2)/(x^2-x-42))÷((5x)/(x^2-36))

Related Questions

Support LCMGCF.com by disabling your adblocker.