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