Multiply $ \dfrac{4x^3-x^2-3x}{x^2-10x+25} $ by $ \dfrac{x^2+3x-10}{x^2-x} $ to get $ \dfrac{4x^3+15x^2-31x-30}{x^2-10x+25} $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{4x^3-x^2-3x}{x^2-10x+25} \cdot \frac{x^2+3x-10}{x^2-x} & \xlongequal{\text{Step 1}} \frac{ \left( 4x+3 \right) \cdot \color{blue}{ \left( x^2-x \right) } }{ x^2-10x+25 } \cdot \frac{ x^2+3x-10 }{ 1 \cdot \color{blue}{ \left( x^2-x \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 4x+3 }{ x^2-10x+25 } \cdot \frac{ x^2+3x-10 }{ 1 } \xlongequal{\text{Step 3}} \frac{ \left( 4x+3 \right) \cdot \left( x^2+3x-10 \right) }{ \left( x^2-10x+25 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 4x^3+12x^2-40x+3x^2+9x-30 }{ x^2-10x+25 } = \frac{4x^3+15x^2-31x-30}{x^2-10x+25} \end{aligned} $$