Divide $ \dfrac{3k^2-28k+60}{5k^2-9k-2} $ by $ \dfrac{3k-10}{5k+1} $ to get $ \dfrac{ k-6 }{ k-2 } $.

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.

$$ \begin{aligned} \frac{ \frac{3k^2-28k+60}{5k^2-9k-2} }{ \frac{\color{blue}{3k-10}}{\color{blue}{5k+1}} } & \xlongequal{\text{Step 1}} \frac{3k^2-28k+60}{5k^2-9k-2} \cdot \frac{\color{blue}{5k+1}}{\color{blue}{3k-10}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( k-6 \right) \cdot \color{blue}{ \left( 3k-10 \right) } }{ \left( k-2 \right) \cdot \color{red}{ \left( 5k+1 \right) } } \cdot \frac{ 1 \cdot \color{red}{ \left( 5k+1 \right) } }{ 1 \cdot \color{blue}{ \left( 3k-10 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ k-6 }{ k-2 } \cdot \frac{ 1 }{ 1 } \xlongequal{\text{Step 4}} \frac{ \left( k-6 \right) \cdot 1 }{ \left( k-2 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 5}} \frac{ k-6 }{ k-2 } \end{aligned} $$