Divide $ \dfrac{4x^3-4x^2-3x}{4x+20} $ by $ x $ to get $ \dfrac{ 4x^3-4x^2-3x }{ 4x^2+20x } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{4x^3-4x^2-3x}{4x+20} }{x} & \xlongequal{\text{Step 1}} \frac{4x^3-4x^2-3x}{4x+20} \cdot \frac{\color{blue}{1}}{\color{blue}{x}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 4x^3-4x^2-3x \right) \cdot 1 }{ \left( 4x+20 \right) \cdot x } \xlongequal{\text{Step 3}} \frac{ 4x^3-4x^2-3x }{ 4x^2+20x } \end{aligned} $$