Factor (x^2-9)/(x^2+5x+6)

$\frac{x^{2} - 9}{x^{2} + 5 x + 6}$

Step 1

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

$\frac{x^{2} - 3^{2}}{x^{2} + 5 x + 6}$

Step 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 $b = 3$ .

$\frac{(x + 3) (x - 3)}{x^{2} + 5 x + 6}$

Step 3

Factor $x^{2} + 5 x + 6$ 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 $6$ and whose sum is $5$ .

$2, 3$

Write the factored form using these integers.

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

Step 4

Reduce the expression $\frac{(x + 3) (x - 3)}{(x + 2) (x + 3)}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

$\frac{x - 3}{x + 2}$

Factor (x^2-9)/(x^2+5x+6)

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