Question
Answer
$$1-(1-x)^4=-x^4+4x^3-6x^2+4x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}1-(1-x)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}1-(x^4-4x^3+6x^2-4x+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}1-x^4+4x^3-6x^2+4x-1 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{1}-x^4+4x^3-6x^2+4x -\cancel{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-x^4+4x^3-6x^2+4x\end{aligned} $$ | |
| ① | $$ (1-x)^4 = (1-x)^2 \cdot (1-x)^2 $$ |
| ② | Find $ \left(1-x\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ x }$. $$ \begin{aligned}\left(1-x\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot x + \color{red}{x^2} = 1-2x+x^2\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{1-2x+x^2}\right) $ by each term in $ \left( 1-2x+x^2\right) $. $$ \left( \color{blue}{1-2x+x^2}\right) \cdot \left( 1-2x+x^2\right) = 1-2x+x^2-2x+4x^2-2x^3+x^2-2x^3+x^4 $$ |
| ④ | Combine like terms: $$ 1 \color{blue}{-2x} + \color{red}{x^2} \color{blue}{-2x} + \color{green}{4x^2} \color{orange}{-2x^3} + \color{green}{x^2} \color{orange}{-2x^3} +x^4 = x^4 \color{orange}{-4x^3} + \color{green}{6x^2} \color{blue}{-4x} +1 $$ |
| ⑤ | Remove the parentheses by changing the sign of each term within them. $$ - \left( x^4-4x^3+6x^2-4x+1 \right) = -x^4+4x^3-6x^2+4x-1 $$ |
| ⑥ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,-x^4+4x^3-6x^2+4x \, \color{blue}{ -\cancel{1}} \, = -x^4+4x^3-6x^2+4x $$ |
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