Answer
$$(x-3)^2-3(x-3)-8=x^2-9x+10$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-3)^2-3(x-3)-8& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2-6x+9-3(x-3)-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2-6x+9-(3x-9)-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^2-6x+9-3x+9-8 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^2-9x+10\end{aligned} $$ | |
| 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-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: $$ x^2 \color{blue}{-6x} + \color{red}{9} \color{blue}{-3x} + \color{green}{9} \color{green}{-8} = x^2 \color{blue}{-9x} + \color{green}{10} $$ |
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