Question
Answer
$$\dfrac{2x^2+20x+48}{2(x^2-3x-28)}=\dfrac{x+6}{x-7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2x^2+20x+48}{2(x^2-3x-28)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^2+20x+48}{2x^2-6x-56} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x+6}{x-7}\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{2} $ by $ \left( x^2-3x-28\right) $ $$ \color{blue}{2} \cdot \left( x^2-3x-28\right) = 2x^2-6x-56 $$ |
| ② | Simplify $ \dfrac{2x^2+20x+48}{2x^2-6x-56} $ to $ \dfrac{x+6}{x-7} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x+8}$. $$ \begin{aligned} \frac{2x^2+20x+48}{2x^2-6x-56} & =\frac{ \left( x+6 \right) \cdot \color{blue}{ \left( 2x+8 \right) }}{ \left( x-7 \right) \cdot \color{blue}{ \left( 2x+8 \right) }} = \\[1ex] &= \frac{x+6}{x-7} \end{aligned} $$ |
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