Question
Answer
$$\dfrac{q^2+3q-28}{q^2+10q+21}=\dfrac{q-4}{q+3}$$
Explanation
| $$ \begin{aligned}\frac{q^2+3q-28}{q^2+10q+21}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{q-4}{q+3}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{q^2+3q-28}{q^2+10q+21} $ to $ \dfrac{q-4}{q+3} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{q+7}$. $$ \begin{aligned} \frac{q^2+3q-28}{q^2+10q+21} & =\frac{ \left( q-4 \right) \cdot \color{blue}{ \left( q+7 \right) }}{ \left( q+3 \right) \cdot \color{blue}{ \left( q+7 \right) }} = \\[1ex] &= \frac{q-4}{q+3} \end{aligned} $$ |
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