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