Answer
$$\frac{9x^2+81x}{x^3+8x^2-9x}=\frac{9}{x-1}$$
Explanation
| $$ \begin{aligned}\frac{9x^2+81x}{x^3+8x^2-9x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{x-1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{9x^2+81x}{x^3+8x^2-9x} $ to $ \dfrac{9}{x-1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2+9x}$. $$ \begin{aligned} \frac{9x^2+81x}{x^3+8x^2-9x} & =\frac{ 9 \cdot \color{blue}{ \left( x^2+9x \right) }}{ \left( x-1 \right) \cdot \color{blue}{ \left( x^2+9x \right) }} = \\[1ex] &= \frac{9}{x-1} \end{aligned} $$ |
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Simplify Rational Expressions