Multiply $ \dfrac{a^2+11a+24}{ab+3b} $ by $ \dfrac{b}{a+8} $ to get $ \dfrac{ ab+3b }{ ab+3b } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{a^2+11a+24}{ab+3b} \cdot \frac{b}{a+8} & \xlongequal{\text{Step 1}} \frac{ \left( a+3 \right) \cdot \color{blue}{ \left( a+8 \right) } }{ ab+3b } \cdot \frac{ b }{ 1 \cdot \color{blue}{ \left( a+8 \right) } } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ a+3 }{ ab+3b } \cdot \frac{ b }{ 1 } \xlongequal{\text{Step 3}} \frac{ \left( a+3 \right) \cdot b }{ \left( ab+3b \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ ab+3b }{ ab+3b } \end{aligned} $$

Simplify $ \dfrac{a+3}{a+3} $ to $ 1$.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{a+3}$.

$$ \begin{aligned} \frac{a+3}{a+3} & =\frac{ 1 \cdot \color{blue}{ \left( a+3 \right) }}{ 1 \cdot \color{blue}{ \left( a+3 \right) }} = \\[1ex] &= \frac{1}{1} =1 \end{aligned} $$