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