Question
Answer
$$(n-3)^4=n^4-12n^3+54n^2-108n+81$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(n-3)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}n^4-12n^3+54n^2-108n+81\end{aligned} $$ | |
| ① | $$ (n-3)^4 = (n-3)^2 \cdot (n-3)^2 $$ |
| ② | Find $ \left(n-3\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}{ 3 }$. $$ \begin{aligned}\left(n-3\right)^2 = \color{blue}{n^2} -2 \cdot n \cdot 3 + \color{red}{3^2} = n^2-6n+9\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{n^2-6n+9}\right) $ by each term in $ \left( n^2-6n+9\right) $. $$ \left( \color{blue}{n^2-6n+9}\right) \cdot \left( n^2-6n+9\right) = n^4-6n^3+9n^2-6n^3+36n^2-54n+9n^2-54n+81 $$ |
| ④ | Combine like terms: $$ n^4 \color{blue}{-6n^3} + \color{red}{9n^2} \color{blue}{-6n^3} + \color{green}{36n^2} \color{orange}{-54n} + \color{green}{9n^2} \color{orange}{-54n} +81 = \\ = n^4 \color{blue}{-12n^3} + \color{green}{54n^2} \color{orange}{-108n} +81 $$ |
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