Simplify denominator

$$ 2x^2+ \color{blue}{12} + \color{blue}{16} = 2x^2+ \color{blue}{28} $$

Subtract $ \dfrac{x+2}{2x^2+28} $ from $ \dfrac{4x+1}{3x+12} $ to get $ \dfrac{ \color{purple}{ 8x^3-x^2+94x+4 } }{ 6x^3+24x^2+84x+336 }$.

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}{ 2x^2+28 }$ and the second by $\color{blue}{ 3x+12 }$.

$$ \begin{aligned} \frac{4x+1}{3x+12} - \frac{x+2}{2x^2+28} & = \frac{ \left( 4x+1 \right) \cdot \color{blue}{ \left( 2x^2+28 \right) }}{ \left( 3x+12 \right) \cdot \color{blue}{ \left( 2x^2+28 \right) }} - \frac{ \left( x+2 \right) \cdot \color{blue}{ \left( 3x+12 \right) }}{ \left( 2x^2+28 \right) \cdot \color{blue}{ \left( 3x+12 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 8x^3+112x+2x^2+28 } }{ 6x^3+84x+24x^2+336 } - \frac{ \color{purple}{ 3x^2+12x+6x+24 } }{ 6x^3+84x+24x^2+336 } = \\[1ex] &=\frac{ \color{purple}{ 8x^3+112x+2x^2+28 - \left( 3x^2+12x+6x+24 \right) } }{ 6x^3+24x^2+84x+336 }=\frac{ \color{purple}{ 8x^3-x^2+94x+4 } }{ 6x^3+24x^2+84x+336 } \end{aligned} $$