Answer
$$\frac{b^2+14b+48}{b^2+2b-24}=\frac{b+8}{b-4}$$
Explanation
| $$ \begin{aligned}\frac{b^2+14b+48}{b^2+2b-24}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{b+8}{b-4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{b^2+14b+48}{b^2+2b-24} $ to $ \dfrac{b+8}{b-4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b+6}$. $$ \begin{aligned} \frac{b^2+14b+48}{b^2+2b-24} & =\frac{ \left( b+8 \right) \cdot \color{blue}{ \left( b+6 \right) }}{ \left( b-4 \right) \cdot \color{blue}{ \left( b+6 \right) }} = \\[1ex] &= \frac{b+8}{b-4} \end{aligned} $$ |
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Simplify Rational Expressions