Question
Answer
$$\dfrac{x^2+x-2}{x(x^2+4x+4)}=\dfrac{x-1}{x^2+2x}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^2+x-2}{x(x^2+4x+4)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^2+x-2}{x^3+4x^2+4x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x-1}{x^2+2x}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{x} $ by $ \left( x^2+4x+4\right) $ $$ \color{blue}{x} \cdot \left( x^2+4x+4\right) = x^3+4x^2+4x $$ |
| ② | Simplify $ \dfrac{x^2+x-2}{x^3+4x^2+4x} $ to $ \dfrac{x-1}{x^2+2x} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+2}$. $$ \begin{aligned} \frac{x^2+x-2}{x^3+4x^2+4x} & =\frac{ \left( x-1 \right) \cdot \color{blue}{ \left( x+2 \right) }}{ \left( x^2+2x \right) \cdot \color{blue}{ \left( x+2 \right) }} = \\[1ex] &= \frac{x-1}{x^2+2x} \end{aligned} $$ |
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Simplify Rational Expressions