Answer
$$\frac{x^3-2x^2-7x-4}{x^2+2x+1}=x-4$$
Explanation
| $$ \begin{aligned}\frac{x^3-2x^2-7x-4}{x^2+2x+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x-4\end{aligned} $$ | |
| ① | Simplify $ \dfrac{x^3-2x^2-7x-4}{x^2+2x+1} $ to $ x-4$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x^2+2x+1}$. $$ \begin{aligned} \frac{x^3-2x^2-7x-4}{x^2+2x+1} & =\frac{ \left( x-4 \right) \cdot \color{blue}{ \left( x^2+2x+1 \right) }}{ 1 \cdot \color{blue}{ \left( x^2+2x+1 \right) }} = \\[1ex] &= \frac{x-4}{1} =x-4 \end{aligned} $$ |
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