Answer
$$\frac{2-18x^2}{1+6x+9x^2}=\frac{-6x+2}{3x+1}$$
Explanation
| $$ \begin{aligned}\frac{2-18x^2}{1+6x+9x^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-6x+2}{3x+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2-18x^2}{1+6x+9x^2} $ to $ \dfrac{-6x+2}{3x+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{3x+1}$. $$ \begin{aligned} \frac{2-18x^2}{1+6x+9x^2} & =\frac{ \left( -6x+2 \right) \cdot \color{blue}{ \left( 3x+1 \right) }}{ \left( 3x+1 \right) \cdot \color{blue}{ \left( 3x+1 \right) }} = \\[1ex] &= \frac{-6x+2}{3x+1} \end{aligned} $$ |
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Simplify Rational Expressions