Answer
$$\frac{d^3+21d^2-46d}{d^2-4}=\frac{d^2+23d}{d+2}$$
Explanation
| $$ \begin{aligned}\frac{d^3+21d^2-46d}{d^2-4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{d^2+23d}{d+2}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{d^3+21d^2-46d}{d^2-4} $ to $ \dfrac{d^2+23d}{d+2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{d-2}$. $$ \begin{aligned} \frac{d^3+21d^2-46d}{d^2-4} & =\frac{ \left( d^2+23d \right) \cdot \color{blue}{ \left( d-2 \right) }}{ \left( d+2 \right) \cdot \color{blue}{ \left( d-2 \right) }} = \\[1ex] &= \frac{d^2+23d}{d+2} \end{aligned} $$ |
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Simplify Rational Expressions