Question
Answer
$$(-x^2+2x)^2(x-1)=x^5-5x^4+8x^3-4x^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-x^2+2x)^2(x-1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x^4-4x^3+4x^2)(x-1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}x^5-x^4-4x^4+4x^3+4x^3-4x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}x^5-5x^4+8x^3-4x^2\end{aligned} $$ | |
| ① | Find $ \left(-x^2+2x\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply formula $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x^2 } $ and $ B = \color{red}{ 2x }$. $$ \begin{aligned}\left(-x^2+2x\right)^2& \xlongequal{ S1 } \left(x^2-2x\right)^2 \xlongequal{ S2 } \color{blue}{\left( x^2 \right)^2} -2 \cdot x^2 \cdot 2x + \color{red}{\left( 2x \right)^2} = \\[1 em] & = x^4-4x^3+4x^2\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{x^4-4x^3+4x^2}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{x^4-4x^3+4x^2}\right) \cdot \left( x-1\right) = x^5-x^4-4x^4+4x^3+4x^3-4x^2 $$ |
| ③ | Combine like terms: $$ x^5 \color{blue}{-x^4} \color{blue}{-4x^4} + \color{red}{4x^3} + \color{red}{4x^3} -4x^2 = x^5 \color{blue}{-5x^4} + \color{red}{8x^3} -4x^2 $$ |
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