Multiply $3$ by $ \dfrac{z}{z^2+9z+20} $ to get $ \dfrac{ 3z }{ z^2+9z+20 } $.

Step 1: Write $ 3 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 3 \cdot \frac{z}{z^2+9z+20} & \xlongequal{\text{Step 1}} \frac{3}{\color{red}{1}} \cdot \frac{z}{z^2+9z+20} \xlongequal{\text{Step 2}} \frac{ 3 \cdot z }{ 1 \cdot \left( z^2+9z+20 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 3z }{ z^2+9z+20 } \end{aligned} $$

Multiply $6$ by $ \dfrac{z}{z^2+10z+24} $ to get $ \dfrac{ 6z }{ z^2+10z+24 } $.

Step 1: Write $ 6 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 6 \cdot \frac{z}{z^2+10z+24} & \xlongequal{\text{Step 1}} \frac{6}{\color{red}{1}} \cdot \frac{z}{z^2+10z+24} \xlongequal{\text{Step 2}} \frac{ 6 \cdot z }{ 1 \cdot \left( z^2+10z+24 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 6z }{ z^2+10z+24 } \end{aligned} $$

Subtract $ \dfrac{6z}{z^2+10z+24} $ from $ \dfrac{3z}{z^2+9z+20} $ to get $ \dfrac{ \color{purple}{ -3z^2-12z } }{ z^3+15z^2+74z+120 }$.

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}{ z+6 }$ and the second by $\color{blue}{ z+5 }$.

$$ \begin{aligned} \frac{3z}{z^2+9z+20} - \frac{6z}{z^2+10z+24} & = \frac{ 3z \cdot \color{blue}{ \left( z+6 \right) }}{ \left( z^2+9z+20 \right) \cdot \color{blue}{ \left( z+6 \right) }} - \frac{ 6z \cdot \color{blue}{ \left( z+5 \right) }}{ \left( z^2+10z+24 \right) \cdot \color{blue}{ \left( z+5 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3z^2+18z } }{ z^3+6z^2+9z^2+54z+20z+120 } - \frac{ \color{purple}{ 6z^2+30z } }{ z^3+6z^2+9z^2+54z+20z+120 } = \\[1ex] &=\frac{ \color{purple}{ 3z^2+18z - \left( 6z^2+30z \right) } }{ z^3+15z^2+74z+120 }=\frac{ \color{purple}{ -3z^2-12z } }{ z^3+15z^2+74z+120 } \end{aligned} $$

Simplify $ \dfrac{-3z^2-12z}{z^3+15z^2+74z+120} $ to $ \dfrac{-3z}{z^2+11z+30} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{z+4}$.

$$ \begin{aligned} \frac{-3z^2-12z}{z^3+15z^2+74z+120} & =\frac{ \left( -3z \right) \cdot \color{blue}{ \left( z+4 \right) }}{ \left( z^2+11z+30 \right) \cdot \color{blue}{ \left( z+4 \right) }} = \\[1ex] &= \frac{-3z}{z^2+11z+30} \end{aligned} $$