Simplify (x^2+2x-3)/(x^2-9)

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

Step 1

Factor $x^{2} + 2 x - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$\frac{(x - 1) (x + 3)}{x^{2} - 9}$

Step 2

Simplify the denominator.

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

$\frac{(x - 1) (x + 3)}{x^{2} - 3^{2}}$

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 $b = 3$ .

$\frac{(x - 1) (x + 3)}{(x + 3) (x - 3)}$

Step 3

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$\frac{(x - 1) (x + 3)}{(x + 3) (x - 3)}$

Rewrite the expression.

$\frac{x - 1}{x - 3}$

Simplify (x^2+2x-3)/(x^2-9)

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