Answer
$$\frac{2x^2-4x}{2x^3-8x}=\frac{1}{x+2}$$
Explanation
| $$ \begin{aligned}\frac{2x^2-4x}{2x^3-8x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x+2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^2-4x}{2x^3-8x} $ to $ \dfrac{1}{x+2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x^2-4x}$. $$ \begin{aligned} \frac{2x^2-4x}{2x^3-8x} & =\frac{ 1 \cdot \color{blue}{ \left( 2x^2-4x \right) }}{ \left( x+2 \right) \cdot \color{blue}{ \left( 2x^2-4x \right) }} = \\[1ex] &= \frac{1}{x+2} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions