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