Question
Answer
$$\dfrac{2x^2+4x-16}{-10+3x+x^2}=\dfrac{2x+8}{x+5}$$
Explanation
| $$ \begin{aligned}\frac{2x^2+4x-16}{-10+3x+x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x+8}{x+5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^2+4x-16}{-10+3x+x^2} $ to $ \dfrac{2x+8}{x+5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-2}$. $$ \begin{aligned} \frac{2x^2+4x-16}{-10+3x+x^2} & =\frac{ \left( 2x+8 \right) \cdot \color{blue}{ \left( x-2 \right) }}{ \left( x+5 \right) \cdot \color{blue}{ \left( x-2 \right) }} = \\[1ex] &= \frac{2x+8}{x+5} \end{aligned} $$ |
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