Answer
$\dfrac{12x}{x + 3}; \quad x \ne -3, 2, 3$
Work Step by Step
First, factor all expressions in the original exercise:
$\dfrac{3x(x - 3)}{x - 2} \div \dfrac{(x - 3)(x + 3)}{4(x - 2)}$
To divide one fraction by another, multiply the first fraction by the reciprocal of the second fraction:
$\dfrac{3x(x - 3)}{x - 2} \cdot \dfrac{4(x - 2)}{(x - 3)(x + 3)}$
Multiply numerator with numerator and denominator with denominator:
$\dfrac{3x(x - 3) \cdot 4(x - 2)}{(x - 2)(x - 3)(x + 3)}$
Cancel common terms in the numerator and denominator:
$\require{cancel}\dfrac{3x\cancel{(x - 3)} \cdot 4\cancel{(x - 2)}}{\cancel{(x - 2)}\cancel{(x - 3)}(x + 3)}=\dfrac{3x \cdot 4}{x + 3}$
Multiply to simplify:
$\dfrac{12x}{x + 3}$
Restrictions on $x$ occur when the value of $x$ makes the fraction undefined, which means that the denominator becomes $0$.
Set denominators equal to $0$ and solve for $x$ to find where the restrictions are:
First denominator:
$x - 2 = 0$
$x = 2$
Second denominator:
$4x - 8 = 0$
$4x = 8$
$x = 2$
Third denominator (from the reciprocal):
$x - 3 = 0$
$x = 3$
Fourth denominator (from the reciprocal):
$x + 3 = 0$
$x = -3$
Restrictions: $x \ne -3, 2, 3$
