Question
Answer
$$\dfrac{b^2-2b-8}{b^2+10b+16}=\dfrac{b-4}{b+8}$$
Explanation
| $$ \begin{aligned}\frac{b^2-2b-8}{b^2+10b+16}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{b-4}{b+8}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{b^2-2b-8}{b^2+10b+16} $ to $ \dfrac{b-4}{b+8} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b+2}$. $$ \begin{aligned} \frac{b^2-2b-8}{b^2+10b+16} & =\frac{ \left( b-4 \right) \cdot \color{blue}{ \left( b+2 \right) }}{ \left( b+8 \right) \cdot \color{blue}{ \left( b+2 \right) }} = \\[1ex] &= \frac{b-4}{b+8} \end{aligned} $$ |
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Simplify Rational Expressions