Answer
$$(x+1)^2-3(x+1)+6=x^2-x+4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+1)^2-3(x+1)+6& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+2x+1-3(x+1)+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+2x+1-(3x+3)+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^2+2x+1-3x-3+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^2-x+4\end{aligned} $$ | |
| 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 | Multiply $ \color{blue}{3} $ by $ \left( x+1\right) $ $$ \color{blue}{3} \cdot \left( x+1\right) = 3x+3 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x+3 \right) = -3x-3 $$ |
| ④ | Combine like terms: $$ x^2+ \color{blue}{2x} + \color{red}{1} \color{blue}{-3x} \color{green}{-3} + \color{green}{6} = x^2 \color{blue}{-x} + \color{green}{4} $$ |
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