Question
Answer
$$\dfrac{80y+80xy+20x^2y}{20y-5x^2y}=\dfrac{4x+8}{-x+2}$$
Explanation
| $$ \begin{aligned}\frac{80y+80xy+20x^2y}{20y-5x^2y}& \xlongequal{ }\frac{16+16x+4x^2}{4-x^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4x+8}{-x+2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{16+16x+4x^2}{4-x^2} $ to $ \dfrac{4x+8}{-x+2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$. $$ \begin{aligned} \frac{16+16x+4x^2}{4-x^2} & =\frac{ \left( 4x+8 \right) \cdot \color{blue}{ \left( x+2 \right) }}{ \left( -x+2 \right) \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] &= \frac{4x+8}{-x+2} \end{aligned} $$ |
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