Question
Answer
$$\dfrac{2x^3+x^2-18x-9}{3x-x^2}=\dfrac{2x^2+7x+3}{-x}$$
Explanation
| $$ \begin{aligned}\frac{2x^3+x^2-18x-9}{3x-x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^2+7x+3}{-x}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2x^3+x^2-18x-9}{3x-x^2} $ to $ \dfrac{2x^2+7x+3}{-x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-3}$. $$ \begin{aligned} \frac{2x^3+x^2-18x-9}{3x-x^2} & =\frac{ \left( 2x^2+7x+3 \right) \cdot \color{blue}{ \left( x-3 \right) }}{ \left( -x \right) \cdot \color{blue}{ \left( x-3 \right) }} = \\[1ex] &= \frac{2x^2+7x+3}{-x} \end{aligned} $$ |
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