Question
Answer
$$\dfrac{2x^2+8x}{16-x^2}=\dfrac{2x}{-x+4}$$
Explanation
| $$ \begin{aligned}\frac{2x^2+8x}{16-x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x}{-x+4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^2+8x}{16-x^2} $ to $ \dfrac{2x}{-x+4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$. $$ \begin{aligned} \frac{2x^2+8x}{16-x^2} & =\frac{ 2x \cdot \color{blue}{ \left( x+4 \right) }}{ \left( -x+4 \right) \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{2x}{-x+4} \end{aligned} $$ |
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Simplify Rational Expressions