Answer
$$\frac{x^2+16x+64}{x^2-2x-80}=\frac{x+8}{x-10}$$
Explanation
| $$ \begin{aligned}\frac{x^2+16x+64}{x^2-2x-80}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x+8}{x-10}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+16x+64}{x^2-2x-80} $ to $ \dfrac{x+8}{x-10} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+8}$. $$ \begin{aligned} \frac{x^2+16x+64}{x^2-2x-80} & =\frac{ \left( x+8 \right) \cdot \color{blue}{ \left( x+8 \right) }}{ \left( x-10 \right) \cdot \color{blue}{ \left( x+8 \right) }} = \\[1ex] &= \frac{x+8}{x-10} \end{aligned} $$ |
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Simplify Rational Expressions