Simplify (x^2+18x+81)/(x^2-81)

$\frac{x^{2} + 18 x + 81}{x^{2} - 81}$

Step 1

Factor using the perfect square rule.

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

$\frac{x^{2} + 18 x + 9^{2}}{x^{2} - 81}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 x = 2 \cdot x \cdot 9$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 9$ .

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

Step 2

Simplify the denominator.

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

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

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 $b = 9$ .

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

Step 3

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

Factor $x + 9$ out of ${(x + 9)}^{2}$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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