Answer
$$4(x+1)^2-(3x+2)(4x+1)=-8x^2-3x+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}4(x+1)^2-(3x+2)(4x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4(x^2+2x+1)-(3x+2)(4x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^2+8x+4-(12x^2+3x+8x+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^2+8x+4-(12x^2+11x+2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4x^2+8x+4-12x^2-11x-2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-8x^2-3x+2\end{aligned} $$ | |
| 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}{4} $ by $ \left( x^2+2x+1\right) $ $$ \color{blue}{4} \cdot \left( x^2+2x+1\right) = 4x^2+8x+4 $$ Multiply each term of $ \left( \color{blue}{3x+2}\right) $ by each term in $ \left( 4x+1\right) $. $$ \left( \color{blue}{3x+2}\right) \cdot \left( 4x+1\right) = 12x^2+3x+8x+2 $$ |
| ③ | Combine like terms: $$ 12x^2+ \color{blue}{3x} + \color{blue}{8x} +2 = 12x^2+ \color{blue}{11x} +2 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 12x^2+11x+2 \right) = -12x^2-11x-2 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{4x^2} + \color{red}{8x} + \color{green}{4} \color{blue}{-12x^2} \color{red}{-11x} \color{green}{-2} = \color{blue}{-8x^2} \color{red}{-3x} + \color{green}{2} $$ |
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