Answer
$$(x-1)(x-2)^2=x^3-5x^2+8x-4$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-1)(x-2)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-1)(x^2-4x+4) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^3-4x^2+4x-x^2+4x-4 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^3-5x^2+8x-4\end{aligned} $$ | |
| 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) $. $$ \left( \color{blue}{x-1}\right) \cdot \left( x^2-4x+4\right) = x^3-4x^2+4x-x^2+4x-4 $$ |
| ③ | Combine like terms: $$ x^3 \color{blue}{-4x^2} + \color{red}{4x} \color{blue}{-x^2} + \color{red}{4x} -4 = x^3 \color{blue}{-5x^2} + \color{red}{8x} -4 $$ |
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