Answer
$$\frac{4x^2+15x+9}{x^2-3x-18}=\frac{4x+3}{x-6}$$
Explanation
| $$ \begin{aligned}\frac{4x^2+15x+9}{x^2-3x-18}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4x+3}{x-6}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{4x^2+15x+9}{x^2-3x-18} $ to $ \dfrac{4x+3}{x-6} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$. $$ \begin{aligned} \frac{4x^2+15x+9}{x^2-3x-18} & =\frac{ \left( 4x+3 \right) \cdot \color{blue}{ \left( x+3 \right) }}{ \left( x-6 \right) \cdot \color{blue}{ \left( x+3 \right) }} = \\[1ex] &= \frac{4x+3}{x-6} \end{aligned} $$ |
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Simplify Rational Expressions