Divide ((x^2-1)/(9x))÷((x^2+7x-8)/(4x))

$\frac{x^{2} - 1}{9 x} \div \frac{x^{2} + 7 x - 8}{4 x}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 1}{9 x} \cdot \frac{4 x}{x^{2} + 7 x - 8}$

Step 2

Cancel the common factor of $x$ .

Factor $x$ out of $9 x$ .

$\frac{x^{2} - 1}{x \cdot 9} \cdot \frac{4 x}{x^{2} + 7 x - 8}$

Factor $x$ out of $4 x$ .

$\frac{x^{2} - 1}{x \cdot 9} \cdot \frac{x \cdot 4}{x^{2} + 7 x - 8}$

Cancel the common factor.

$\frac{x^{2} - 1}{x \cdot 9} \cdot \frac{x \cdot 4}{x^{2} + 7 x - 8}$

Rewrite the expression.

$\frac{x^{2} - 1}{9} \cdot \frac{4}{x^{2} + 7 x - 8}$

Step 3

Multiply $\frac{x^{2} - 1}{9}$ by $\frac{4}{x^{2} + 7 x - 8}$ .

$\frac{(x^{2} - 1) \cdot 4}{9 (x^{2} + 7 x - 8)}$

Step 4

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$\frac{(x^{2} - 1^{2}) \cdot 4}{9 (x^{2} + 7 x - 8)}$

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 $b = 1$ .

$\frac{(x + 1) (x - 1) \cdot 4}{9 (x^{2} + 7 x - 8)}$

Step 5

Factor $x^{2} + 7 x - 8$ 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 $- 8$ and whose sum is $7$ .

$- 1, 8$

Write the factored form using these integers.

$\frac{(x + 1) (x - 1) \cdot 4}{9 ((x - 1) (x + 8))}$

$\frac{(x + 1) (x - 1) \cdot 4}{9 (x - 1) (x + 8)}$

Step 6

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{(x + 1) (x - 1) \cdot 4}{9 (x - 1) (x + 8)}$

Rewrite the expression.

$\frac{(x + 1) \cdot 4}{9 (x + 8)}$

Step 7

Move $4$ to the left of $x + 1$ .

$\frac{4 (x + 1)}{9 (x + 8)}$

Divide ((x^2-1)/(9x))÷((x^2+7x-8)/(4x))

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