Answer
$$2(x+3)\cdot3(x-2)^2=6x^3-6x^2-48x+72$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2(x+3)\cdot3(x-2)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2(x+3)\cdot3(x^2-4x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(2x+6)\cdot3(x^2-4x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(6x+18)(x^2-4x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}6x^3-24x^2+24x+18x^2-72x+72 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}6x^3-6x^2-48x+72\end{aligned} $$ | |
| ① | Find $ \left(x-2\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}{ 2 }$. $$ \begin{aligned}\left(x-2\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 2 + \color{red}{2^2} = x^2-4x+4\end{aligned} $$ |
| ② | Multiply $ \color{blue}{2} $ by $ \left( x+3\right) $ $$ \color{blue}{2} \cdot \left( x+3\right) = 2x+6 $$ |
| ③ | $$ \left( \color{blue}{2x+6}\right) \cdot 3 = 6x+18 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{6x+18}\right) $ by each term in $ \left( x^2-4x+4\right) $. $$ \left( \color{blue}{6x+18}\right) \cdot \left( x^2-4x+4\right) = 6x^3-24x^2+24x+18x^2-72x+72 $$ |
| ⑤ | Combine like terms: $$ 6x^3 \color{blue}{-24x^2} + \color{red}{24x} + \color{blue}{18x^2} \color{red}{-72x} +72 = 6x^3 \color{blue}{-6x^2} \color{red}{-48x} +72 $$ |
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