Answer
$\dfrac{4}{x}, x \ne -5, 0, 1, 4$
Work Step by Step
To divide one rational expression by another, change the operation to multiplication and invert the divisor:
$\dfrac{6x^3 - 6x^2}{x^4 + 5x^3} \cdot \dfrac{2x^2 + 2x - 40}{3x^2 - 15x + 12}$
Factor each polynomial completely:
$=\dfrac{6x^2(x - 1)}{x^3(x + 5)} \cdot \dfrac{2(x^2 + x - 20)}{3(x^2 - 5x + 4)}$
$=\dfrac{6x^2(x - 1)}{x^3(x + 5)} \cdot \dfrac{2(x+ 5)(x - 4)}{3(x - 4)(x - 1)}$
Cancel common factors in the numerator and denominator:
$\require{cancel}=\dfrac{\cancel{6}^2\cancel{x^2}\cancel{(x - 1)}}{x^{\cancel{3}}\cancel{(x + 5)}} \cdot \dfrac{2\cancel{(x+ 5)}\cancel{(x - 4)}}{\cancel{3}\cancel{(x - 4)}\cancel{(x - 1)}}$
$=\dfrac{2 \cdot 2}{x}$
$=\dfrac{4}{x}$
Restrictions occur when the denominator is $0$, meaning the expression becomes undefined. To find the restrictions, set each expression in the denominators of the original rational expressions and in the reciprocal equal to zero and solve:
Restriction:
$x^3 = 0$
$x = 0$
Restriction:
$x + 5 = 0$
$x = -5$
Restriction:
$x - 4 = 0$
$x = 4$
Restriction:
$x - 1 = 0$
$x = 1$
Therefore, $x \ne 0, -5, 4, \text{ or } 1$.
