Subtract $2x$ from $ \dfrac{7}{x^2} $ to get $ \dfrac{ \color{purple}{ -2x^3+7 } }{ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{7}{x^2} -2x & \xlongequal{\text{Step 1}} \frac{7}{x^2} - \frac{2x}{\color{red}{1}} = \frac{ 7 }{ x^2 } - \frac{ 2x \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 7 } }{ x^2 } - \frac{ \color{purple}{ 2x^3 } }{ x^2 }=\frac{ \color{purple}{ -2x^3+7 } }{ x^2 } \end{aligned} $$

Subtract $15$ from $ \dfrac{-2x^3+7}{x^2} $ to get $ \dfrac{ \color{purple}{ -2x^3-15x^2+7 } }{ x^2 }$.

Step 1: Write $ 15 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{-2x^3+7}{x^2} -15 & \xlongequal{\text{Step 1}} \frac{-2x^3+7}{x^2} - \frac{15}{\color{red}{1}} = \frac{ -2x^3+7 }{ x^2 } - \frac{ 15 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -2x^3+7 } }{ x^2 } - \frac{ \color{purple}{ 15x^2 } }{ x^2 }=\frac{ \color{purple}{ -2x^3-15x^2+7 } }{ x^2 } \end{aligned} $$