Question
Answer
$$\dfrac{4x^2-36x}{2x^4-24x^3+54x^2}=\dfrac{2}{x^2-3x}$$
Explanation
| $$ \begin{aligned}\frac{4x^2-36x}{2x^4-24x^3+54x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2}{x^2-3x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^2-36x}{2x^4-24x^3+54x^2} $ to $ \dfrac{2}{x^2-3x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x^2-18x}$. $$ \begin{aligned} \frac{4x^2-36x}{2x^4-24x^3+54x^2} & =\frac{ 2 \cdot \color{blue}{ \left( 2x^2-18x \right) }}{ \left( x^2-3x \right) \cdot \color{blue}{ \left( 2x^2-18x \right) }} = \\[1ex] &= \frac{2}{x^2-3x} \end{aligned} $$ |
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