Multiply $8$ by $ \dfrac{x}{3} $ to get $ \dfrac{ 8x }{ 3 } $.

Step 1: Write $ 8 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 8 \cdot \frac{x}{3} & \xlongequal{\text{Step 1}} \frac{8}{\color{red}{1}} \cdot \frac{x}{3} \xlongequal{\text{Step 2}} \frac{ 8 \cdot x }{ 1 \cdot 3 } \xlongequal{\text{Step 3}} \frac{ 8x }{ 3 } \end{aligned} $$

Subtract $ \dfrac{8x}{3} $ from $ \dfrac{5}{2x^2+10x} $ to get $ \dfrac{ \color{purple}{ -16x^3-80x^2+15 } }{ 6x^2+30x }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 3 }$ and the second by $\color{blue}{ 2x^2+10x }$.

$$ \begin{aligned} \frac{5}{2x^2+10x} - \frac{8x}{3} & = \frac{ 5 \cdot \color{blue}{ 3 }}{ \left( 2x^2+10x \right) \cdot \color{blue}{ 3 }} - \frac{ 8x \cdot \color{blue}{ \left( 2x^2+10x \right) }}{ 3 \cdot \color{blue}{ \left( 2x^2+10x \right) }} = \\[1ex] &=\frac{ \color{purple}{ 15 } }{ 6x^2+30x } - \frac{ \color{purple}{ 16x^3+80x^2 } }{ 6x^2+30x }=\frac{ \color{purple}{ 15 - \left( 16x^3+80x^2 \right) } }{ 6x^2+30x } = \\[1ex] &=\frac{ \color{purple}{ -16x^3-80x^2+15 } }{ 6x^2+30x } \end{aligned} $$