# How do you simplify the expression (x^2 + x - 6)/(x^2 - 4) * (x^2 - 9)/( x^2 + 6x + 9)?

Mar 11, 2018

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

#### Explanation:

first, factorise each expression.

$3 + - 2 = 1$
$3 \cdot - 2 = - 6$

${x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

$3 + 3 = 6$
$3 \cdot 3 = 9$

${x}^{2} + 6 x + 9 = \left(x + 3\right) \left(x + 3\right)$

for the other two expressions, the identity $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ can be used.

${x}^{2} - 4 = {x}^{2} - {2}^{2}$
$= \left(x + 2\right) \left(x - 2\right)$

${x}^{2} - 9 = {x}^{2} - {3}^{2}$
$= \left(x + 3\right) \left(x - 3\right)$

putting the factorised expressions into the question gives

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

this can be cancelled

$\frac{\left(x + 3\right) \cancel{\left(x - 2\right)}}{\left(x + 2\right) \cancel{\left(x - 2\right)}} \cdot \frac{\cancel{\left(x + 3\right)} \left(x - 3\right)}{\cancel{\left(x + 3\right)} \left(x + 3\right)}$

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

$\frac{\cancel{\left(x + 3\right)}}{\left(x + 2\right)} \cdot \frac{\left(x - 3\right)}{\cancel{\left(x + 3\right)}}$

to give $\frac{1}{x + 2} \cdot \frac{x - 3}{1}$

this is the same as $\frac{x - 3}{x + 2}$