Question
Answer
$$\dfrac{k^2+13k+40}{k^2-2k-35}=\dfrac{k+8}{k-7}$$
Explanation
| $$ \begin{aligned}\frac{k^2+13k+40}{k^2-2k-35}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{k+8}{k-7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{k^2+13k+40}{k^2-2k-35} $ to $ \dfrac{k+8}{k-7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{k+5}$. $$ \begin{aligned} \frac{k^2+13k+40}{k^2-2k-35} & =\frac{ \left( k+8 \right) \cdot \color{blue}{ \left( k+5 \right) }}{ \left( k-7 \right) \cdot \color{blue}{ \left( k+5 \right) }} = \\[1ex] &= \frac{k+8}{k-7} \end{aligned} $$ |
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Simplify Rational Expressions