Subtract $8x+1$ from $ \dfrac{x}{3x^2+42x-8} $ to get $ \dfrac{ \color{purple}{ -24x^3-339x^2+23x+8 } }{ 3x^2+42x-8 }$.

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

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3x^2+42x-8}$.

$$ \begin{aligned} \frac{x}{3x^2+42x-8} -8x+1 & \xlongequal{\text{Step 1}} \frac{x}{3x^2+42x-8} - \frac{8x+1}{\color{red}{1}} = \frac{ x }{ 3x^2+42x-8 } - \frac{ \left( 8x+1 \right) \cdot \color{blue}{ \left( 3x^2+42x-8 \right) }}{ 1 \cdot \color{blue}{ \left( 3x^2+42x-8 \right) }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x } }{ 3x^2+42x-8 } - \frac{ \color{purple}{ 24x^3+336x^2-64x+3x^2+42x-8 } }{ 3x^2+42x-8 } = \\[1ex] &=\frac{ \color{purple}{ x - \left( 24x^3+336x^2-64x+3x^2+42x-8 \right) } }{ 3x^2+42x-8 }=\frac{ \color{purple}{ -24x^3-339x^2+23x+8 } }{ 3x^2+42x-8 } \end{aligned} $$