Answer
$$\frac{x^3+4x^2-32x}{32x-8x^2}=\frac{x+8}{-8}$$
Explanation
| $$ \begin{aligned}\frac{x^3+4x^2-32x}{32x-8x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+8}{-8}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^3+4x^2-32x}{32x-8x^2} $ to $ \dfrac{x+8}{-8} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2-4x}$. $$ \begin{aligned} \frac{x^3+4x^2-32x}{32x-8x^2} & =\frac{ \left( x+8 \right) \cdot \color{blue}{ \left( x^2-4x \right) }}{ \left( -8 \right) \cdot \color{blue}{ \left( x^2-4x \right) }} = \\[1ex] &= \frac{x+8}{-8} \end{aligned} $$ |
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