Question
Answer
$$\dfrac{4v^3-49v}{8v^2-38v+35}=\dfrac{2v^2+7v}{4v-5}$$
Explanation
| $$ \begin{aligned}\frac{4v^3-49v}{8v^2-38v+35}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2v^2+7v}{4v-5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4v^3-49v}{8v^2-38v+35} $ to $ \dfrac{2v^2+7v}{4v-5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2v-7}$. $$ \begin{aligned} \frac{4v^3-49v}{8v^2-38v+35} & =\frac{ \left( 2v^2+7v \right) \cdot \color{blue}{ \left( 2v-7 \right) }}{ \left( 4v-5 \right) \cdot \color{blue}{ \left( 2v-7 \right) }} = \\[1ex] &= \frac{2v^2+7v}{4v-5} \end{aligned} $$ |
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