Answer
$$3(x+3)^2-3(x+3)-3=3x^2+15x+15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3(x+3)^2-3(x+3)-3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3(x^2+6x+9)-3(x+3)-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^2+18x+27-(3x+9)-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^2+18x+27-3x-9-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3x^2+15x+15\end{aligned} $$ | |
| ① | Find $ \left(x+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}{ x } $ and $ B = \color{red}{ 3 }$. $$ \begin{aligned}\left(x+3\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 3 + \color{red}{3^2} = x^2+6x+9\end{aligned} $$ |
| ② | Multiply $ \color{blue}{3} $ by $ \left( x^2+6x+9\right) $ $$ \color{blue}{3} \cdot \left( x^2+6x+9\right) = 3x^2+18x+27 $$Multiply $ \color{blue}{3} $ by $ \left( x+3\right) $ $$ \color{blue}{3} \cdot \left( x+3\right) = 3x+9 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3x+9 \right) = -3x-9 $$ |
| ④ | Combine like terms: $$ 3x^2+ \color{blue}{18x} + \color{red}{27} \color{blue}{-3x} \color{green}{-9} \color{green}{-3} = 3x^2+ \color{blue}{15x} + \color{green}{15} $$ |
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