Question
Answer
$$(x+3)^2-5(x+3)+1=x^2+x-5$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+3)^2-5(x+3)+1& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}x^2+6x+9-5(x+3)+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^2+6x+9-(5x+15)+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^2+6x+9-5x-15+1 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^2+x-5\end{aligned} $$ | |
| 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}{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: $$ x^2+ \color{blue}{6x} + \color{red}{9} \color{blue}{-5x} \color{green}{-15} + \color{green}{1} = x^2+ \color{blue}{x} \color{green}{-5} $$ |
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