Question
Answer
$$\dfrac{z^2-9z+18}{z^2+z-12}=\dfrac{z-6}{z+4}$$
Explanation
| $$ \begin{aligned}\frac{z^2-9z+18}{z^2+z-12}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{z-6}{z+4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{z^2-9z+18}{z^2+z-12} $ to $ \dfrac{z-6}{z+4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{z-3}$. $$ \begin{aligned} \frac{z^2-9z+18}{z^2+z-12} & =\frac{ \left( z-6 \right) \cdot \color{blue}{ \left( z-3 \right) }}{ \left( z+4 \right) \cdot \color{blue}{ \left( z-3 \right) }} = \\[1ex] &= \frac{z-6}{z+4} \end{aligned} $$ |
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Simplify Rational Expressions