Divide $ \dfrac{2x^2+10}{x^2+6x+5} $ by $ \dfrac{x^2+6x+9}{x^2-x-2} $ to get $ \dfrac{2x^3-4x^2+10x-20}{x^3+11x^2+39x+45} $.

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

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{2x^2+10}{x^2+6x+5} }{ \frac{\color{blue}{x^2+6x+9}}{\color{blue}{x^2-x-2}} } & \xlongequal{\text{Step 1}} \frac{2x^2+10}{x^2+6x+5} \cdot \frac{\color{blue}{x^2-x-2}}{\color{blue}{x^2+6x+9}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 2x^2+10 }{ \left( x+5 \right) \cdot \color{red}{ \left( x+1 \right) } } \cdot \frac{ \left( x-2 \right) \cdot \color{red}{ \left( x+1 \right) } }{ x^2+6x+9 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^2+10 }{ x+5 } \cdot \frac{ x-2 }{ x^2+6x+9 } \xlongequal{\text{Step 4}} \frac{ \left( 2x^2+10 \right) \cdot \left( x-2 \right) }{ \left( x+5 \right) \cdot \left( x^2+6x+9 \right) } = \\[1ex] & \xlongequal{\text{Step 5}} \frac{ 2x^3-4x^2+10x-20 }{ x^3+6x^2+9x+5x^2+30x+45 } = \frac{2x^3-4x^2+10x-20}{x^3+11x^2+39x+45} \end{aligned} $$