Answer
$$\frac{2x^2+8x+8}{(2x+4)^2}=\frac{1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2x^2+8x+8}{(2x+4)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2x^2+8x+8}{4x^2+16x+16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{2}\end{aligned} $$ | |
| ① | Find $ \left(2x+4\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2x } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(2x+4\right)^2 = \color{blue}{\left( 2x \right)^2} +2 \cdot 2x \cdot 4 + \color{red}{4^2} = 4x^2+16x+16\end{aligned} $$ |
| ② | Simplify $ \dfrac{2x^2+8x+8}{4x^2+16x+16} $ to $ \dfrac{1}{2} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x^2+8x+8}$. $$ \begin{aligned} \frac{2x^2+8x+8}{4x^2+16x+16} & =\frac{ 1 \cdot \color{blue}{ \left( 2x^2+8x+8 \right) }}{ 2 \cdot \color{blue}{ \left( 2x^2+8x+8 \right) }} = \\[1ex] &= \frac{1}{2} \end{aligned} $$ |
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