Question
Answer
$$(n-2)^4=n^4-8n^3+24n^2-32n+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-2)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n^4-8n^3+24n^2-32n+16\end{aligned} $$ | |
| ① | $$ (n-2)^4 = (n-2)^2 \cdot (n-2)^2 $$ |
| ② | Find $ \left(n-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}{ n } $ and $ B = \color{red}{ 2 }$. $$ \begin{aligned}\left(n-2\right)^2 = \color{blue}{n^2} -2 \cdot n \cdot 2 + \color{red}{2^2} = n^2-4n+4\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{n^2-4n+4}\right) $ by each term in $ \left( n^2-4n+4\right) $. $$ \left( \color{blue}{n^2-4n+4}\right) \cdot \left( n^2-4n+4\right) = n^4-4n^3+4n^2-4n^3+16n^2-16n+4n^2-16n+16 $$ |
| ④ | Combine like terms: $$ n^4 \color{blue}{-4n^3} + \color{red}{4n^2} \color{blue}{-4n^3} + \color{green}{16n^2} \color{orange}{-16n} + \color{green}{4n^2} \color{orange}{-16n} +16 = \\ = n^4 \color{blue}{-8n^3} + \color{green}{24n^2} \color{orange}{-32n} +16 $$ |
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