Question
Answer
$$(x-1)^2(x+2)(2x+1)=2x^4+x^3-6x^2+x+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-1)^2(x+2)(2x+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^2-2x+1)(x+2)(2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(x^3+2x^2-2x^2-4x+x+2)(2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(x^3-3x+2)(2x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2x^4+x^3-6x^2-3x+4x+2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2x^4+x^3-6x^2+x+2\end{aligned} $$ | |
| 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( x+2\right) $. $$ \left( \color{blue}{x^2-2x+1}\right) \cdot \left( x+2\right) = x^3+ \cancel{2x^2} -\cancel{2x^2}-4x+x+2 $$ |
| ③ | Combine like terms: $$ x^3+ \, \color{blue}{ \cancel{2x^2}} \, \, \color{blue}{ -\cancel{2x^2}} \, \color{green}{-4x} + \color{green}{x} +2 = x^3 \color{green}{-3x} +2 $$ |
| ④ | Multiply each term of $ \left( \color{blue}{x^3-3x+2}\right) $ by each term in $ \left( 2x+1\right) $. $$ \left( \color{blue}{x^3-3x+2}\right) \cdot \left( 2x+1\right) = 2x^4+x^3-6x^2-3x+4x+2 $$ |
| ⑤ | Combine like terms: $$ 2x^4+x^3-6x^2 \color{blue}{-3x} + \color{blue}{4x} +2 = 2x^4+x^3-6x^2+ \color{blue}{x} +2 $$ |
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