Question
Answer
$$\dfrac{2k^3+5k^2-18k}{k-2}=2k^2+9k$$
Explanation
| $$ \begin{aligned}\frac{2k^3+5k^2-18k}{k-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2k^2+9k\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2k^3+5k^2-18k}{k-2} $ to $ 2k^2+9k$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{k-2}$. $$ \begin{aligned} \frac{2k^3+5k^2-18k}{k-2} & =\frac{ \left( 2k^2+9k \right) \cdot \color{blue}{ \left( k-2 \right) }}{ 1 \cdot \color{blue}{ \left( k-2 \right) }} = \\[1ex] &= \frac{2k^2+9k}{1} =2k^2+9k \end{aligned} $$ |
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Simplify Rational Expressions