Simplify $ \dfrac{3x^2-6x-72}{5x^2-35x+30} $ to $ \dfrac{3x+12}{5x-5} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-6}$.

$$ \begin{aligned} \frac{3x^2-6x-72}{5x^2-35x+30} & =\frac{ \left( 3x+12 \right) \cdot \color{blue}{ \left( x-6 \right) }}{ \left( 5x-5 \right) \cdot \color{blue}{ \left( x-6 \right) }} = \\[1ex] &= \frac{3x+12}{5x-5} \end{aligned} $$

Multiply $ \dfrac{3x+12}{5x-5} $ by $ \dfrac{x^2+x-2}{x^2+9x+20} $ to get $ \dfrac{ 3x+6 }{ 5x+25 } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{3x+12}{5x-5} \cdot \frac{x^2+x-2}{x^2+9x+20} & \xlongequal{\text{Step 1}} \frac{ 3 \cdot \color{blue}{ \left( x+4 \right) } }{ 5 \cdot \color{red}{ \left( x-1 \right) } } \cdot \frac{ \left( x+2 \right) \cdot \color{red}{ \left( x-1 \right) } }{ \left( x+5 \right) \cdot \color{blue}{ \left( x+4 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3 }{ 5 } \cdot \frac{ x+2 }{ x+5 } \xlongequal{\text{Step 3}} \frac{ 3 \cdot \left( x+2 \right) }{ 5 \cdot \left( x+5 \right) } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 3x+6 }{ 5x+25 } \end{aligned} $$

Subtract $25$ from $ \dfrac{x}{x^2} $ to get $ \dfrac{ \color{purple}{ -25x^2+x } }{ x^2 }$.

Step 1: Write $ 25 $ 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 $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x}{x^2} -25 & \xlongequal{\text{Step 1}} \frac{x}{x^2} - \frac{25}{\color{red}{1}} = \frac{ x }{ x^2 } - \frac{ 25 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x } }{ x^2 } - \frac{ \color{purple}{ 25x^2 } }{ x^2 }=\frac{ \color{purple}{ -25x^2+x } }{ x^2 } \end{aligned} $$

Subtract $ \dfrac{-25x^2+x}{x^2} $ from $ \dfrac{3x+6}{5x+25} $ to get $ \dfrac{ \color{purple}{ 128x^3+626x^2-25x } }{ 5x^3+25x^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 }$ and the second by $\color{blue}{ 5x+25 }$.

$$ \begin{aligned} \frac{3x+6}{5x+25} - \frac{-25x^2+x}{x^2} & = \frac{ \left( 3x+6 \right) \cdot \color{blue}{ x^2 }}{ \left( 5x+25 \right) \cdot \color{blue}{ x^2 }} - \frac{ \left( -25x^2+x \right) \cdot \color{blue}{ \left( 5x+25 \right) }}{ x^2 \cdot \color{blue}{ \left( 5x+25 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3x^3+6x^2 } }{ 5x^3+25x^2 } - \frac{ \color{purple}{ -125x^3-625x^2+5x^2+25x } }{ 5x^3+25x^2 } = \\[1ex] &=\frac{ \color{purple}{ 3x^3+6x^2 - \left( -125x^3-625x^2+5x^2+25x \right) } }{ 5x^3+25x^2 }=\frac{ \color{purple}{ 128x^3+626x^2-25x } }{ 5x^3+25x^2 } \end{aligned} $$