Question
Answer
$$\dfrac{4x^3+21x^2+27x+28}{x+4}=4x^2+5x+7$$
Explanation
| $$ \begin{aligned}\frac{4x^3+21x^2+27x+28}{x+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2+5x+7\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^3+21x^2+27x+28}{x+4} $ to $ 4x^2+5x+7$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+4}$. $$ \begin{aligned} \frac{4x^3+21x^2+27x+28}{x+4} & =\frac{ \left( 4x^2+5x+7 \right) \cdot \color{blue}{ \left( x+4 \right) }}{ 1 \cdot \color{blue}{ \left( x+4 \right) }} = \\[1ex] &= \frac{4x^2+5x+7}{1} =4x^2+5x+7 \end{aligned} $$ |
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