Question
Answer
$$\dfrac{3b^2-48b+180}{b^2-18b+80}=\dfrac{3b-18}{b-8}$$
Explanation
| $$ \begin{aligned}\frac{3b^2-48b+180}{b^2-18b+80}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3b-18}{b-8}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{3b^2-48b+180}{b^2-18b+80} $ to $ \dfrac{3b-18}{b-8} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b-10}$. $$ \begin{aligned} \frac{3b^2-48b+180}{b^2-18b+80} & =\frac{ \left( 3b-18 \right) \cdot \color{blue}{ \left( b-10 \right) }}{ \left( b-8 \right) \cdot \color{blue}{ \left( b-10 \right) }} = \\[1ex] &= \frac{3b-18}{b-8} \end{aligned} $$ |
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Simplify Rational Expressions