Simplify (x^2-81)/(x^2-6x-27)*(x+3)/x

$\frac{x^{2} - 81}{x^{2} - 6 x - 27} \cdot \frac{x + 3}{x}$

Step 1

Simplify the numerator.

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

$\frac{x^{2} - 9^{2}}{x^{2} - 6 x - 27} \cdot \frac{x + 3}{x}$

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 $b = 9$ .

$\frac{(x + 9) (x - 9)}{x^{2} - 6 x - 27} \cdot \frac{x + 3}{x}$

Step 2

Factor $x^{2} - 6 x - 27$ 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 $- 27$ and whose sum is $- 6$ .

$- 9, 3$

Write the factored form using these integers.

$\frac{(x + 9) (x - 9)}{(x - 9) (x + 3)} \cdot \frac{x + 3}{x}$

Step 3

Simplify terms.

Cancel the common factor of $x + 3$ .

Factor $x + 3$ out of $(x - 9) (x + 3)$ .

$\frac{(x + 9) (x - 9)}{(x + 3) (x - 9)} \cdot \frac{x + 3}{x}$

Cancel the common factor.

$\frac{(x + 9) (x - 9)}{(x + 3) (x - 9)} \cdot \frac{x + 3}{x}$

Rewrite the expression.

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

Multiply $\frac{(x + 9) (x - 9)}{x - 9}$ by $\frac{1}{x}$ .

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

Cancel the common factor of $x - 9$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{x + 9}{x}$

Simplify (x^2-81)/(x^2-6x-27)*(x+3)/x

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