Answer
$$\frac{x^2+5x+4}{x^2-6x-7}=\frac{x+4}{x-7}$$
Explanation
| $$ \begin{aligned}\frac{x^2+5x+4}{x^2-6x-7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+4}{x-7}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+5x+4}{x^2-6x-7} $ to $ \dfrac{x+4}{x-7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+1}$. $$ \begin{aligned} \frac{x^2+5x+4}{x^2-6x-7} & =\frac{ \left( x+4 \right) \cdot \color{blue}{ \left( x+1 \right) }}{ \left( x-7 \right) \cdot \color{blue}{ \left( x+1 \right) }} = \\[1ex] &= \frac{x+4}{x-7} \end{aligned} $$ |
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Simplify Rational Expressions