Question
Answer
$$(k+1)^2(2k^2+4k+1)=2k^4+8k^3+11k^2+6k+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(k+1)^2(2k^2+4k+1)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1k^2+2k+1)(2k^2+4k+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}2k^4+8k^3+11k^2+6k+1\end{aligned} $$ | |
| ① | Find $ \left(k+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}{ k } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(k+1\right)^2 = \color{blue}{k^2} +2 \cdot k \cdot 1 + \color{red}{1^2} = k^2+2k+1\end{aligned} $$ |
| ② | Multiply each term of $ \left( \color{blue}{k^2+2k+1}\right) $ by each term in $ \left( 2k^2+4k+1\right) $. $$ \left( \color{blue}{k^2+2k+1}\right) \cdot \left( 2k^2+4k+1\right) = 2k^4+4k^3+k^2+4k^3+8k^2+2k+2k^2+4k+1 $$ |
| ③ | Combine like terms: $$ 2k^4+ \color{blue}{4k^3} + \color{red}{k^2} + \color{blue}{4k^3} + \color{green}{8k^2} + \color{orange}{2k} + \color{green}{2k^2} + \color{orange}{4k} +1 = \\ = 2k^4+ \color{blue}{8k^3} + \color{green}{11k^2} + \color{orange}{6k} +1 $$ |
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