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