Answer
$$3(x-3)^2-5(x-3)-3=3x^2-23x+39$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}3(x-3)^2-5(x-3)-3& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3(x^2-6x+9)-5(x-3)-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3x^2-18x+27-(5x-15)-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}3x^2-18x+27-5x+15-3 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}3x^2-23x+39\end{aligned} $$ | |
| 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}{5} $ by $ \left( x-3\right) $ $$ \color{blue}{5} \cdot \left( x-3\right) = 5x-15 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 5x-15 \right) = -5x+15 $$ |
| ④ | Combine like terms: $$ 3x^2 \color{blue}{-18x} + \color{red}{27} \color{blue}{-5x} + \color{green}{15} \color{green}{-3} = 3x^2 \color{blue}{-23x} + \color{green}{39} $$ |
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