Simplify ((x^2-16)/(81-x^2)*(2x)/(x^2+10x+24))÷((x-4)/(x^2+15x+54))

$\frac{x^{2} - 16}{81 - x^{2}} \cdot \frac{2 x}{x^{2} + 10 x + 24} \div \frac{x - 4}{x^{2} + 15 x + 54}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 16}{81 - x^{2}} \cdot \frac{2 x}{x^{2} + 10 x + 24} \frac{x^{2} + 15 x + 54}{x - 4}$

Step 2

Simplify the numerator.

Rewrite $16$ as $4^{2}$ .

$\frac{x^{2} - 4^{2}}{81 - x^{2}} \cdot \frac{2 x}{x^{2} + 10 x + 24} \frac{x^{2} + 15 x + 54}{x - 4}$

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 $b = 4$ .

$\frac{(x + 4) (x - 4)}{81 - x^{2}} \cdot \frac{2 x}{x^{2} + 10 x + 24} \frac{x^{2} + 15 x + 54}{x - 4}$

Step 3

Simplify the denominator.

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

$\frac{(x + 4) (x - 4)}{9^{2} - x^{2}} \cdot \frac{2 x}{x^{2} + 10 x + 24} \frac{x^{2} + 15 x + 54}{x - 4}$

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

$\frac{(x + 4) (x - 4)}{(9 + x) (9 - x)} \cdot \frac{2 x}{x^{2} + 10 x + 24} \frac{x^{2} + 15 x + 54}{x - 4}$

Step 4

Factor $x^{2} + 10 x + 24$ 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 $24$ and whose sum is $10$ .

$4, 6$

Write the factored form using these integers.

$\frac{(x + 4) (x - 4)}{(9 + x) (9 - x)} \cdot \frac{2 x}{(x + 4) (x + 6)} \frac{x^{2} + 15 x + 54}{x - 4}$

Step 5

Simplify terms.

Cancel the common factor of $x + 4$ .

Cancel the common factor.

$\frac{(x + 4) (x - 4)}{(9 + x) (9 - x)} \cdot \frac{2 x}{(x + 4) (x + 6)} \frac{x^{2} + 15 x + 54}{x - 4}$

Rewrite the expression.

$\frac{x - 4}{(9 + x) (9 - x)} \cdot \frac{2 x}{x + 6} \frac{x^{2} + 15 x + 54}{x - 4}$

Multiply $\frac{x - 4}{(9 + x) (9 - x)}$ by $\frac{2 x}{x + 6}$ .

$\frac{(x - 4) (2 x)}{(9 + x) (9 - x) (x + 6)} \cdot \frac{x^{2} + 15 x + 54}{x - 4}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{(x - 4) (2 x)}{(9 + x) (9 - x) (x + 6)} \cdot \frac{x^{2} + 15 x + 54}{x - 4}$

Rewrite the expression.

$\frac{2 x}{(9 + x) (9 - x) (x + 6)} (x^{2} + 15 x + 54)$

Multiply $\frac{2 x}{(9 + x) (9 - x) (x + 6)}$ by $x^{2} + 15 x + 54$ .

$\frac{2 x (x^{2} + 15 x + 54)}{(9 + x) (9 - x) (x + 6)}$

Step 6

Factor $x^{2} + 15 x + 54$ 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 $54$ and whose sum is $15$ .

$6, 9$

Write the factored form using these integers.

$\frac{2 x ((x + 6) (x + 9))}{(9 + x) (9 - x) (x + 6)}$

$\frac{2 x (x + 6) (x + 9)}{(9 + x) (9 - x) (x + 6)}$

Step 7

Cancel the common factor of $x + 6$ .

Cancel the common factor.

$\frac{2 x (x + 6) (x + 9)}{(9 + x) (9 - x) (x + 6)}$

Rewrite the expression.

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

Step 8

Cancel the common factor of $x + 9$ and $9 + x$ .

Reorder terms.

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 x}{9 - x}$

Simplify ((x^2-16)/(81-x^2)*(2x)/(x^2+10x+24))÷((x-4)/(x^2+15x+54))

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