Answer
$$\frac{y^2-2y-8}{32-2y^2}=\frac{y+2}{-2y-8}$$
Explanation
| $$ \begin{aligned}\frac{y^2-2y-8}{32-2y^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{y+2}{-2y-8}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{y^2-2y-8}{32-2y^2} $ to $ \dfrac{y+2}{-2y-8} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{y-4}$. $$ \begin{aligned} \frac{y^2-2y-8}{32-2y^2} & =\frac{ \left( y+2 \right) \cdot \color{blue}{ \left( y-4 \right) }}{ \left( -2y-8 \right) \cdot \color{blue}{ \left( y-4 \right) }} = \\[1ex] &= \frac{y+2}{-2y-8} \end{aligned} $$ |
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