Answer
$$(2x+3)^2-(x-1)(x+2)=3x^2+11x+11$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x+3)^2-(x-1)(x+2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2+12x+9-(x-1)(x+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^2+12x+9-(x^2+2x-x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^2+12x+9-(x^2+x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4x^2+12x+9-x^2-x+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}3x^2+11x+11\end{aligned} $$ | |
| ① | Find $ \left(2x+3\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}{ 3 }$. $$ \begin{aligned}\left(2x+3\right)^2 = \color{blue}{\left( 2x \right)^2} +2 \cdot 2x \cdot 3 + \color{red}{3^2} = 4x^2+12x+9\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x+2\right) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x+2\right) = x^2+2x-x-2 $$ |
| ③ | Combine like terms: $$ x^2+ \color{blue}{2x} \color{blue}{-x} -2 = x^2+ \color{blue}{x} -2 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^2+x-2 \right) = -x^2-x+2 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{4x^2} + \color{red}{12x} + \color{green}{9} \color{blue}{-x^2} \color{red}{-x} + \color{green}{2} = \color{blue}{3x^2} + \color{red}{11x} + \color{green}{11} $$ |
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