Question
Answer
$$(2k+1)^4+4(2k+1)^2+11=16k^4+32k^3+40k^2+24k+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2k+1)^4+4(2k+1)^2+11& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16k^4+32k^3+24k^2+8k+1+4(4k^2+4k+1)+11 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}16k^4+32k^3+24k^2+8k+1+16k^2+16k+4+11 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}16k^4+32k^3+40k^2+24k+5+11 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}16k^4+32k^3+40k^2+24k+16\end{aligned} $$ | |
| ① | $$ (2k+1)^4 = (2k+1)^2 \cdot (2k+1)^2 $$ |
| ② | Find $ \left(2k+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}{ 2k } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2k+1\right)^2 = \color{blue}{\left( 2k \right)^2} +2 \cdot 2k \cdot 1 + \color{red}{1^2} = 4k^2+4k+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4k^2+4k+1}\right) $ by each term in $ \left( 4k^2+4k+1\right) $. $$ \left( \color{blue}{4k^2+4k+1}\right) \cdot \left( 4k^2+4k+1\right) = 16k^4+16k^3+4k^2+16k^3+16k^2+4k+4k^2+4k+1 $$ |
| ④ | Combine like terms: $$ 16k^4+ \color{blue}{16k^3} + \color{red}{4k^2} + \color{blue}{16k^3} + \color{green}{16k^2} + \color{orange}{4k} + \color{green}{4k^2} + \color{orange}{4k} +1 = \\ = 16k^4+ \color{blue}{32k^3} + \color{green}{24k^2} + \color{orange}{8k} +1 $$Find $ \left(2k+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}{ 2k } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2k+1\right)^2 = \color{blue}{\left( 2k \right)^2} +2 \cdot 2k \cdot 1 + \color{red}{1^2} = 4k^2+4k+1\end{aligned} $$ |
| ⑤ | Multiply $ \color{blue}{4} $ by $ \left( 4k^2+4k+1\right) $ $$ \color{blue}{4} \cdot \left( 4k^2+4k+1\right) = 16k^2+16k+4 $$ |
| ⑥ | Combine like terms: $$ 16k^4+32k^3+ \color{blue}{24k^2} + \color{red}{8k} + \color{green}{1} + \color{blue}{16k^2} + \color{red}{16k} + \color{green}{4} = \\ = 16k^4+32k^3+ \color{blue}{40k^2} + \color{red}{24k} + \color{green}{5} $$ |
| ⑦ | Combine like terms: $$ 16k^4+32k^3+40k^2+24k+ \color{blue}{5} + \color{blue}{11} = 16k^4+32k^3+40k^2+24k+ \color{blue}{16} $$ |
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