Question
Answer
$$\dfrac{4x^3-4x^2-7x+12}{2x+3}=2x^2-5x+4$$
Explanation
| $$ \begin{aligned}\frac{4x^3-4x^2-7x+12}{2x+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2x^2-5x+4\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^3-4x^2-7x+12}{2x+3} $ to $ 2x^2-5x+4$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x+3}$. $$ \begin{aligned} \frac{4x^3-4x^2-7x+12}{2x+3} & =\frac{ \left( 2x^2-5x+4 \right) \cdot \color{blue}{ \left( 2x+3 \right) }}{ 1 \cdot \color{blue}{ \left( 2x+3 \right) }} = \\[1ex] &= \frac{2x^2-5x+4}{1} =2x^2-5x+4 \end{aligned} $$ |
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