Question
Answer
$$\dfrac{4r^2+24r+32}{16r^2-48r-160}=\dfrac{r+4}{4r-20}$$
Explanation
| $$ \begin{aligned}\frac{4r^2+24r+32}{16r^2-48r-160}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{r+4}{4r-20}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4r^2+24r+32}{16r^2-48r-160} $ to $ \dfrac{r+4}{4r-20} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{4r+8}$. $$ \begin{aligned} \frac{4r^2+24r+32}{16r^2-48r-160} & =\frac{ \left( r+4 \right) \cdot \color{blue}{ \left( 4r+8 \right) }}{ \left( 4r-20 \right) \cdot \color{blue}{ \left( 4r+8 \right) }} = \\[1ex] &= \frac{r+4}{4r-20} \end{aligned} $$ |
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Simplify Rational Expressions