Question
Answer
$$\dfrac{2k^2-19k+24}{4k^3-26k^2+30k}=\dfrac{k-8}{2k^2-10k}$$
Explanation
| $$ \begin{aligned}\frac{2k^2-19k+24}{4k^3-26k^2+30k}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{k-8}{2k^2-10k}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2k^2-19k+24}{4k^3-26k^2+30k} $ to $ \dfrac{k-8}{2k^2-10k} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2k-3}$. $$ \begin{aligned} \frac{2k^2-19k+24}{4k^3-26k^2+30k} & =\frac{ \left( k-8 \right) \cdot \color{blue}{ \left( 2k-3 \right) }}{ \left( 2k^2-10k \right) \cdot \color{blue}{ \left( 2k-3 \right) }} = \\[1ex] &= \frac{k-8}{2k^2-10k} \end{aligned} $$ |
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Simplify Rational Expressions