Answer
$$-\frac{1}{(x+1)^2\cdot\frac{2}{(x+1)^{0.5}}}=\frac{1}{2x^2+4x+2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{1}{(x+1)^2\cdot\frac{2}{(x+1)^{0.5}}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{(x+1)^2\cdot\frac{2}{1}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{(x+1)^2\cdot2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{(x^2+2x+1)\cdot2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{1}{2x^2+4x+2}\end{aligned} $$ | |
| ① | A non-zero polynomial raised to the power of 0 equals 1. |
| ② | Remove 1 from denominator. |
| ③ | Find $ \left(x+1\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ④ | $$ \left( \color{blue}{x^2+2x+1}\right) \cdot 2 = 2x^2+4x+2 $$ |
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