Question
Answer
$$\dfrac{2x^3-18x}{2x^2-8x+6}=\dfrac{x^2+3x}{x-1}$$
Explanation
| $$ \begin{aligned}\frac{2x^3-18x}{2x^2-8x+6}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+3x}{x-1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^3-18x}{2x^2-8x+6} $ to $ \dfrac{x^2+3x}{x-1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-6}$. $$ \begin{aligned} \frac{2x^3-18x}{2x^2-8x+6} & =\frac{ \left( x^2+3x \right) \cdot \color{blue}{ \left( 2x-6 \right) }}{ \left( x-1 \right) \cdot \color{blue}{ \left( 2x-6 \right) }} = \\[1ex] &= \frac{x^2+3x}{x-1} \end{aligned} $$ |
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Simplify Rational Expressions