Answer
$$(2x-3)^2-3(x-2)(x+1)=x^2-9x+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2x-3)^2-3(x-2)(x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4x^2-12x+9-3(x-2)(x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x^2-12x+9-(3x-6)(x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4x^2-12x+9-(3x^2+3x-6x-6) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4x^2-12x+9-(3x^2-3x-6) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}4x^2-12x+9-3x^2+3x+6 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}x^2-9x+15\end{aligned} $$ | |
| 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 $ \color{blue}{3} $ by $ \left( x-2\right) $ $$ \color{blue}{3} \cdot \left( x-2\right) = 3x-6 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{3x-6}\right) $ by each term in $ \left( x+1\right) $. $$ \left( \color{blue}{3x-6}\right) \cdot \left( x+1\right) = 3x^2+3x-6x-6 $$ |
| ④ | Combine like terms: $$ 3x^2+ \color{blue}{3x} \color{blue}{-6x} -6 = 3x^2 \color{blue}{-3x} -6 $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x^2-3x-6 \right) = -3x^2+3x+6 $$ |
| ⑥ | Combine like terms: $$ \color{blue}{4x^2} \color{red}{-12x} + \color{green}{9} \color{blue}{-3x^2} + \color{red}{3x} + \color{green}{6} = \color{blue}{x^2} \color{red}{-9x} + \color{green}{15} $$ |
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