Question
Answer
$$\dfrac{x^3-9x}{3x^2-6x-9}=\dfrac{x^2+3x}{3x+3}$$
Explanation
| $$ \begin{aligned}\frac{x^3-9x}{3x^2-6x-9}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+3x}{3x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^3-9x}{3x^2-6x-9} $ to $ \dfrac{x^2+3x}{3x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-3}$. $$ \begin{aligned} \frac{x^3-9x}{3x^2-6x-9} & =\frac{ \left( x^2+3x \right) \cdot \color{blue}{ \left( x-3 \right) }}{ \left( 3x+3 \right) \cdot \color{blue}{ \left( x-3 \right) }} = \\[1ex] &= \frac{x^2+3x}{3x+3} \end{aligned} $$ |
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