Question
Answer
$$\dfrac{n^2+8n+12}{n^2+2n-24}=\dfrac{n+2}{n-4}$$
Explanation
| $$ \begin{aligned}\frac{n^2+8n+12}{n^2+2n-24}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{n+2}{n-4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{n^2+8n+12}{n^2+2n-24} $ to $ \dfrac{n+2}{n-4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{n+6}$. $$ \begin{aligned} \frac{n^2+8n+12}{n^2+2n-24} & =\frac{ \left( n+2 \right) \cdot \color{blue}{ \left( n+6 \right) }}{ \left( n-4 \right) \cdot \color{blue}{ \left( n+6 \right) }} = \\[1ex] &= \frac{n+2}{n-4} \end{aligned} $$ |
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Simplify Rational Expressions