Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-1)(x+2)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-1)(x^2+4x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3+4x^2+4x-x^2-4x-4 \xlongequal{ } \\[1 em] & \xlongequal{ }x^3+4x^2+ \cancel{4x}-x^2 -\cancel{4x}-4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^3+3x^2-4\end{aligned} $$ | |
| 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 each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x^2+4x+4\right) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x^2+4x+4\right) = x^3+4x^2+ \cancel{4x}-x^2 -\cancel{4x}-4 $$ |
| ③ | Combine like terms: $$ x^3+ \color{blue}{4x^2} + \, \color{red}{ \cancel{4x}} \, \color{blue}{-x^2} \, \color{red}{ -\cancel{4x}} \,-4 = x^3+ \color{blue}{3x^2} -4 $$ |