Simplify (x^2+6x+9)/(x^2-9)

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

Step 1

Factor using the perfect square rule.

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

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

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

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

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

Step 2

Simplify the denominator.

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

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

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 $b = 3$ .

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

Step 3

Cancel the common factor of ${(x + 3)}^{2}$ and $x + 3$ .

Factor $x + 3$ out of ${(x + 3)}^{2}$ .

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

Cancel the common factors.

Cancel the common factor.

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

Rewrite the expression.

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