Simplify ((x^4-81)/(x^2+9))÷((x^2+2x-15)/(x^2+5x))

$\frac{x^{4} - 81}{x^{2} + 9} \div \frac{x^{2} + 2 x - 15}{x^{2} + 5 x}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{4} - 81}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Step 2

Simplify the numerator.

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$\frac{{(x^{2})}^{2} - 81}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Rewrite $81$ as $9^{2}$ .

$\frac{{(x^{2})}^{2} - 9^{2}}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 9$ .

$\frac{(x^{2} + 9) (x^{2} - 9)}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Simplify.

Rewrite $9$ as $3^{2}$ .

$\frac{(x^{2} + 9) (x^{2} - 3^{2})}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

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 $b = 3$ .

$\frac{(x^{2} + 9) (x + 3) (x - 3)}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Step 3

Cancel the common factor of $x^{2} + 9$ .

Cancel the common factor.

$\frac{(x^{2} + 9) (x + 3) (x - 3)}{x^{2} + 9} \cdot \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Divide $(x + 3) (x - 3)$ by $1$ .

$(x + 3) (x - 3) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Step 4

Expand $(x + 3) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$(x (x - 3) + 3 (x - 3)) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Apply the distributive property.

$(x \cdot x + x \cdot - 3 + 3 (x - 3)) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Apply the distributive property.

$(x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Step 5

Simplify terms.

Combine the opposite terms in $x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3$ .

Reorder the factors in the terms $x \cdot - 3$ and $3 x$ .

$(x \cdot x - 3 x + 3 x + 3 \cdot - 3) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Add $- 3 x$ and $3 x$ .

$(x \cdot x + 0 + 3 \cdot - 3) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

Add $x \cdot x$ and $0$ .

$(x \cdot x + 3 \cdot - 3) \frac{x^{2} + 5 x}{x^{2} + 2 x - 15}$

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

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

$(x \cdot x + 3 \cdot - 3) \frac{x \cdot x + 5 x}{x^{2} + 2 x - 15}$

Factor $x$ out of $5 x$ .

$(x \cdot x + 3 \cdot - 3) \frac{x \cdot x + x \cdot 5}{x^{2} + 2 x - 15}$

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

$(x \cdot x + 3 \cdot - 3) \frac{x (x + 5)}{x^{2} + 2 x - 15}$

Step 6

Factor $x^{2} + 2 x - 15$ 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 $- 15$ and whose sum is $2$ .

$- 3, 5$

Write the factored form using these integers.

$(x \cdot x + 3 \cdot - 3) \frac{x (x + 5)}{(x - 3) (x + 5)}$

Step 7

Simplify terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Multiply $3$ by $- 3$ .

$(x^{2} - 9) \frac{x (x + 5)}{(x - 3) (x + 5)}$

Simplify terms.

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$(x^{2} - 9) \frac{x (x + 5)}{(x - 3) (x + 5)}$

Rewrite the expression.

$(x^{2} - 9) \frac{x}{x - 3}$

Multiply $x^{2} - 9$ by $\frac{x}{x - 3}$ .

$\frac{(x^{2} - 9) x}{x - 3}$

Step 8

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

$\frac{(x^{2} - 3^{2}) x}{x - 3}$

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 $b = 3$ .

$\frac{(x + 3) (x - 3) x}{x - 3}$

Step 9

Simplify terms.

Cancel the common factor of $x - 3$ .

Cancel the common factor.

$\frac{(x + 3) (x - 3) x}{x - 3}$

Divide $(x + 3) x$ by $1$ .

$(x + 3) x$

Apply the distributive property.

$x \cdot x + 3 x$

Multiply $x$ by $x$ .

$x^{2} + 3 x$

Simplify ((x^4-81)/(x^2+9))÷((x^2+2x-15)/(x^2+5x))

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