Answer
$$\frac{x^3-x^2-12x}{x^3-9x}=\frac{x-4}{x-3}$$
Explanation
| $$ \begin{aligned}\frac{x^3-x^2-12x}{x^3-9x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-4}{x-3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^3-x^2-12x}{x^3-9x} $ to $ \dfrac{x-4}{x-3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2+3x}$. $$ \begin{aligned} \frac{x^3-x^2-12x}{x^3-9x} & =\frac{ \left( x-4 \right) \cdot \color{blue}{ \left( x^2+3x \right) }}{ \left( x-3 \right) \cdot \color{blue}{ \left( x^2+3x \right) }} = \\[1ex] &= \frac{x-4}{x-3} \end{aligned} $$ |
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