Answer
$$(2x-1)^2-(3x+2)(3x-2)=-5x^2-4x+5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x-1)^2-(3x+2)(3x-2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2-4x+1-(3x+2)(3x-2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^2-4x+1-(9x^2-6x+6x-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^2-4x+1-(9x^2-4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4x^2-4x+1-9x^2+4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-5x^2-4x+5\end{aligned} $$ | |
| ① | Find $ \left(2x-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}{ 2x } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2x-1\right)^2 = \color{blue}{\left( 2x \right)^2} -2 \cdot 2x \cdot 1 + \color{red}{1^2} = 4x^2-4x+1\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{3x+2}\right) $ by each term in $ \left( 3x-2\right) $. $$ \left( \color{blue}{3x+2}\right) \cdot \left( 3x-2\right) = 9x^2 -\cancel{6x}+ \cancel{6x}-4 $$ |
| ③ | Combine like terms: $$ 9x^2 \, \color{blue}{ -\cancel{6x}} \,+ \, \color{blue}{ \cancel{6x}} \,-4 = 9x^2-4 $$ |
| ④ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 9x^2-4 \right) = -9x^2+4 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{4x^2} -4x+ \color{red}{1} \color{blue}{-9x^2} + \color{red}{4} = \color{blue}{-5x^2} -4x+ \color{red}{5} $$ |
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