Question
Answer
$$\dfrac{x^2+8x+16}{x+4}=x+4$$
Explanation
| $$ \begin{aligned}\frac{x^2+8x+16}{x+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x+4\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^2+8x+16}{x+4} $ to $ x+4$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$. $$ \begin{aligned} \frac{x^2+8x+16}{x+4} & =\frac{ \left( x+4 \right) \cdot \color{blue}{ \left( x+4 \right) }}{ 1 \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{x+4}{1} =x+4 \end{aligned} $$ |
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