Question
Answer
$$\dfrac{-4x+8}{2x^3+10x^2-28x}=-\dfrac{2}{x^2+7x}$$
Explanation
| $$ \begin{aligned}\frac{-4x+8}{2x^3+10x^2-28x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{2}{x^2+7x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{-4x+8}{2x^3+10x^2-28x} $ to $ \dfrac{-2}{x^2+7x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-4}$. $$ \begin{aligned} \frac{-4x+8}{2x^3+10x^2-28x} & =\frac{ \left( -2 \right) \cdot \color{blue}{ \left( 2x-4 \right) }}{ \left( x^2+7x \right) \cdot \color{blue}{ \left( 2x-4 \right) }} = \\[1ex] &= \frac{-2}{x^2+7x} \end{aligned} $$ |
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