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

Step 1: Write $ x^2+6x $ 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 $\color{blue}{ x^2 }$.

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

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

Step 1: Write $ 2x $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^4+6x^3-5}{x^2} -2x & \xlongequal{\text{Step 1}} \frac{x^4+6x^3-5}{x^2} - \frac{2x}{\color{red}{1}} = \frac{ x^4+6x^3-5 }{ x^2 } - \frac{ 2x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^4+6x^3-5 } }{ x^2 } - \frac{ \color{purple}{ 2x^3 } }{ x^2 }=\frac{ \color{purple}{ x^4+4x^3-5 } }{ x^2 } \end{aligned} $$