Question
Answer
$$\dfrac{x^2+x-72}{2x^4+16x^3-18x^2}=\dfrac{x-8}{2x^3-2x^2}$$
Explanation
| $$ \begin{aligned}\frac{x^2+x-72}{2x^4+16x^3-18x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x-8}{2x^3-2x^2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+x-72}{2x^4+16x^3-18x^2} $ to $ \dfrac{x-8}{2x^3-2x^2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+9}$. $$ \begin{aligned} \frac{x^2+x-72}{2x^4+16x^3-18x^2} & =\frac{ \left( x-8 \right) \cdot \color{blue}{ \left( x+9 \right) }}{ \left( 2x^3-2x^2 \right) \cdot \color{blue}{ \left( x+9 \right) }} = \\[1ex] &= \frac{x-8}{2x^3-2x^2} \end{aligned} $$ |
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