Question
Answer
$$\dfrac{3w^2-48}{w^2+9w+20}=\dfrac{3w-12}{w+5}$$
Explanation
| $$ \begin{aligned}\frac{3w^2-48}{w^2+9w+20}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3w-12}{w+5}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{3w^2-48}{w^2+9w+20} $ to $ \dfrac{3w-12}{w+5} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{w+4}$. $$ \begin{aligned} \frac{3w^2-48}{w^2+9w+20} & =\frac{ \left( 3w-12 \right) \cdot \color{blue}{ \left( w+4 \right) }}{ \left( w+5 \right) \cdot \color{blue}{ \left( w+4 \right) }} = \\[1ex] &= \frac{3w-12}{w+5} \end{aligned} $$ |
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Simplify Rational Expressions