Question
Answer
$$\dfrac{c^2-3c-40}{c^2+12c+35}=\dfrac{c-8}{c+7}$$
Explanation
| $$ \begin{aligned}\frac{c^2-3c-40}{c^2+12c+35}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{c-8}{c+7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{c^2-3c-40}{c^2+12c+35} $ to $ \dfrac{c-8}{c+7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{c+5}$. $$ \begin{aligned} \frac{c^2-3c-40}{c^2+12c+35} & =\frac{ \left( c-8 \right) \cdot \color{blue}{ \left( c+5 \right) }}{ \left( c+7 \right) \cdot \color{blue}{ \left( c+5 \right) }} = \\[1ex] &= \frac{c-8}{c+7} \end{aligned} $$ |
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Simplify Rational Expressions