Question
Answer
$$\dfrac{2x^3+16x^2+24x}{x^2-x-6}=\dfrac{2x^2+12x}{x-3}$$
Explanation
| $$ \begin{aligned}\frac{2x^3+16x^2+24x}{x^2-x-6}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^2+12x}{x-3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^3+16x^2+24x}{x^2-x-6} $ to $ \dfrac{2x^2+12x}{x-3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$. $$ \begin{aligned} \frac{2x^3+16x^2+24x}{x^2-x-6} & =\frac{ \left( 2x^2+12x \right) \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x-3 \right) \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] &= \frac{2x^2+12x}{x-3} \end{aligned} $$ |
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