Simplify $ \dfrac{2y^2+5y-12}{4y^2-8y+3} $ to $ \dfrac{y+4}{2y-1} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2y-3}$.

$$ \begin{aligned} \frac{2y^2+5y-12}{4y^2-8y+3} & =\frac{ \left( y+4 \right) \cdot \color{blue}{ \left( 2y-3 \right) }}{ \left( 2y-1 \right) \cdot \color{blue}{ \left( 2y-3 \right) }} = \\[1ex] &= \frac{y+4}{2y-1} \end{aligned} $$

Divide $ \dfrac{y+4}{2y-1} $ by $ \dfrac{y^2+8y+16}{2y-1} $ to get $ \dfrac{ 1 }{ y+4 } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Cancel $ \color{red}{ 2y-1 } $ in first and second fraction.

Step 3: Factor numerators and denominators.

Step 4: Cancel common factors.

Step 5: Multiply numerators and denominators.

Step 6: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{y+4}{2y-1} }{ \frac{\color{blue}{y^2+8y+16}}{\color{blue}{2y-1}} } & \xlongequal{\text{Step 1}} \frac{y+4}{2y-1} \cdot \frac{\color{blue}{2y-1}}{\color{blue}{y^2+8y+16}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{y+4}{\color{red}{1}} \cdot \frac{\color{red}{1}}{y^2+8y+16} \xlongequal{\text{Step 3}} \frac{ 1 \cdot \color{blue}{ \left( y+4 \right) } }{ 1 } \cdot \frac{ 1 }{ \left( y+4 \right) \cdot \color{blue}{ \left( y+4 \right) } } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 1 }{ 1 } \cdot \frac{ 1 }{ y+4 } \xlongequal{\text{Step 5}} \frac{ 1 \cdot 1 }{ 1 \cdot \left( y+4 \right) } \xlongequal{\text{Step 6}} \frac{ 1 }{ y+4 } \end{aligned} $$