Answer
$$\frac{2x^2+x}{2x^3+7x^2+3x}=\frac{1}{x+3}$$
Explanation
| $$ \begin{aligned}\frac{2x^2+x}{2x^3+7x^2+3x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{x+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^2+x}{2x^3+7x^2+3x} $ to $ \dfrac{1}{x+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x^2+x}$. $$ \begin{aligned} \frac{2x^2+x}{2x^3+7x^2+3x} & =\frac{ 1 \cdot \color{blue}{ \left( 2x^2+x \right) }}{ \left( x+3 \right) \cdot \color{blue}{ \left( 2x^2+x \right) }} = \\[1ex] &= \frac{1}{x+3} \end{aligned} $$ |
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Simplify Rational Expressions