Question
Answer
$$\dfrac{6x^4+7x^3+8x^2+5x-12}{3x^2+2x-3}=2x^2+x+4$$
Explanation
| $$ \begin{aligned}\frac{6x^4+7x^3+8x^2+5x-12}{3x^2+2x-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2x^2+x+4\end{aligned} $$ | |
| ① | Simplify $ \dfrac{6x^4+7x^3+8x^2+5x-12}{3x^2+2x-3} $ to $ 2x^2+x+4$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{3x^2+2x-3}$. $$ \begin{aligned} \frac{6x^4+7x^3+8x^2+5x-12}{3x^2+2x-3} & =\frac{ \left( 2x^2+x+4 \right) \cdot \color{blue}{ \left( 3x^2+2x-3 \right) }}{ 1 \cdot \color{blue}{ \left( 3x^2+2x-3 \right) }} = \\[1ex] &= \frac{2x^2+x+4}{1} =2x^2+x+4 \end{aligned} $$ |
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