Answer
$$\frac{6x^3-96x}{4x^2-64}=\frac{3x}{2}$$
Explanation
| $$ \begin{aligned}\frac{6x^3-96x}{4x^2-64}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3x}{2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{6x^3-96x}{4x^2-64} $ to $ \dfrac{3x}{2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x^2-32}$. $$ \begin{aligned} \frac{6x^3-96x}{4x^2-64} & =\frac{ 3x \cdot \color{blue}{ \left( 2x^2-32 \right) }}{ 2 \cdot \color{blue}{ \left( 2x^2-32 \right) }} = \\[1ex] &= \frac{3x}{2} \end{aligned} $$ |
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