Answer
$$(x+1)^2-x-9=x^2+x-8$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+1)^2-x-9& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+2x+1-x-9 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+x-8\end{aligned} $$ | |
| ① | 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 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ② | Combine like terms: $$ x^2+ \color{blue}{2x} + \color{red}{1} \color{blue}{-x} \color{red}{-9} = x^2+ \color{blue}{x} \color{red}{-8} $$ |
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