Question
Answer
$$\dfrac{5z-5}{2z^3-21z^2+19z}=\dfrac{5}{2z^2-19z}$$
Explanation
| $$ \begin{aligned}\frac{5z-5}{2z^3-21z^2+19z}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{2z^2-19z}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{5z-5}{2z^3-21z^2+19z} $ to $ \dfrac{5}{2z^2-19z} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{z-1}$. $$ \begin{aligned} \frac{5z-5}{2z^3-21z^2+19z} & =\frac{ 5 \cdot \color{blue}{ \left( z-1 \right) }}{ \left( 2z^2-19z \right) \cdot \color{blue}{ \left( z-1 \right) }} = \\[1ex] &= \frac{5}{2z^2-19z} \end{aligned} $$ |
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Simplify Rational Expressions