Answer
$$(1-x^2)^4=x^8-4x^6+6x^4-4x^2+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-x^2)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}x^8-4x^6+6x^4-4x^2+1\end{aligned} $$ | |
| ① | $$ (1-x^2)^4 = (1-x^2)^2 \cdot (1-x^2)^2 $$ |
| ② | Find $ \left(1-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}{ 1 } $ and $ B = \color{red}{ x^2 }$. $$ \begin{aligned}\left(1-x^2\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot x^2 + \color{red}{\left( x^2 \right)^2} = 1-2x^2+x^4\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{1-2x^2+x^4}\right) $ by each term in $ \left( 1-2x^2+x^4\right) $. $$ \left( \color{blue}{1-2x^2+x^4}\right) \cdot \left( 1-2x^2+x^4\right) = 1-2x^2+x^4-2x^2+4x^4-2x^6+x^4-2x^6+x^8 $$ |
| ④ | Combine like terms: $$ 1 \color{blue}{-2x^2} + \color{red}{x^4} \color{blue}{-2x^2} + \color{green}{4x^4} \color{orange}{-2x^6} + \color{green}{x^4} \color{orange}{-2x^6} +x^8 = \\ = x^8 \color{orange}{-4x^6} + \color{green}{6x^4} \color{blue}{-4x^2} +1 $$ |
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