Subtract $ \dfrac{37}{x+5} $ from $ x^3+6x^2+12x $ to get $ \dfrac{ \color{purple}{ x^4+11x^3+42x^2+60x-37 } }{ x+5 }$.

Step 1: Write $ x^3+6x^2+12x $ 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 first fraction by $\color{blue}{ x+5 }$.

$$ \begin{aligned} x^3+6x^2+12x- \frac{37}{x+5} & \xlongequal{\text{Step 1}} \frac{x^3+6x^2+12x}{\color{red}{1}} - \frac{37}{x+5} = \frac{ \left( x^3+6x^2+12x \right) \cdot \color{blue}{ \left( x+5 \right) }}{ 1 \cdot \color{blue}{ \left( x+5 \right) }} - \frac{ 37 }{ x+5 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+5x^3+6x^3+30x^2+12x^2+60x } }{ x+5 } - \frac{ \color{purple}{ 37 } }{ x+5 } = \\[1ex] &=\frac{ \color{purple}{ x^4+11x^3+42x^2+60x-37 } }{ x+5 } \end{aligned} $$