Answer
$\dfrac{18x}{(x + 9)(x + 3)}$
Restriction: $x \ne -9, -3, 3$
Work Step by Step
Let's factor out common factors first:
$\dfrac{2x(x - 3)}{x^2 + 18x + 81} \cdot \dfrac{9(x + 9)}{x^2 - 9}$
Now we can factor out polynomials:
$x^2 + 18x + 81$ = $(x + 9)(x + 9)$
$x^2 - 9$ = $(x + 3)(x - 3)$
Multiply the two expressions using the factors instead of the polynomials:
$\dfrac{2x(x - 3)9(x + 9)}{(x + 9)(x + 9)(x + 3)(x - 3)}$
Factor out $x - 3$, $x + 9$:
$\dfrac{2x(9)}{(x + 9)(x + 3)}$
Multiply out the numerical coefficients:
$\dfrac{18x}{(x + 9)(x + 3)}$
To check what restrictions we have for the variables, we need to find which values of the variables will make the denominators of the original expressions equal $0$, which would make the fraction undefined. Let's set the denominators equal to $0$, and then solve:
First denominator:
$(x + 9)(x + 9) = 0$
Set $x + 9$ equal to zero:
First factor:
$x + 9 = 0$
Subtract $9$ from each side:
$x = -9$
Second denominator:
$(x + 3)(x - 3) = 0$
Set each factor equal to zero:
First factor:
$x + 3 = 0$
Subtract $3$ from each side:
$x = -3$
Second factor:
$x - 3 = 0$
Add $3$ to each side of the equation:
$x = 3$
Restrictions: $x \ne -9, -3, 3$
