Answer
$$\frac{3a^2+21a-90}{3a^2+31a+10}=\frac{3a-9}{3a+1}$$
Explanation
| $$ \begin{aligned}\frac{3a^2+21a-90}{3a^2+31a+10}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3a-9}{3a+1}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{3a^2+21a-90}{3a^2+31a+10} $ to $ \dfrac{3a-9}{3a+1} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{a+10}$. $$ \begin{aligned} \frac{3a^2+21a-90}{3a^2+31a+10} & =\frac{ \left( 3a-9 \right) \cdot \color{blue}{ \left( a+10 \right) }}{ \left( 3a+1 \right) \cdot \color{blue}{ \left( a+10 \right) }} = \\[1ex] &= \frac{3a-9}{3a+1} \end{aligned} $$ |
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Simplify Rational Expressions