Question
Answer
$$(x+1)^2(x-0)(x-1)=x^4+x^3-x^2-x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x+1)^2(x-0)(x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2+2x+1)(x-0)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^3+0x^2+2x^2+0x+x+0)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^3+2x^2+x)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^4-x^3+2x^3-2x^2+x^2-x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}x^4+x^3-x^2-x\end{aligned} $$ | |
| ① | Find $ \left(x+1\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}{ 1 }$. $$ \begin{aligned}\left(x+1\right)^2 = \color{blue}{x^2} +2 \cdot x \cdot 1 + \color{red}{1^2} = x^2+2x+1\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^2+2x+1}\right) $ by each term in $ \left( x0\right) $. $$ \left( \color{blue}{x^2+2x+1}\right) \cdot \left( x0\right) = x^30x^2+2x^20x+x0 $$ |
| ③ | Combine like terms: $$ x^3 \color{blue}{0x^2} + \color{blue}{2x^2} \color{red}{0x} + \color{red}{x} 0 = x^3+ \color{blue}{2x^2} + \color{red}{x} $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^3+2x^2+x}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{x^3+2x^2+x}\right) \cdot \left( x-1\right) = x^4-x^3+2x^3-2x^2+x^2-x $$ |
| ⑤ | Combine like terms: $$ x^4 \color{blue}{-x^3} + \color{blue}{2x^3} \color{red}{-2x^2} + \color{red}{x^2} -x = x^4+ \color{blue}{x^3} \color{red}{-x^2} -x $$ |
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