Question
Answer
$$(2t+1)^4-(2t-1)^4=64t^3+16t$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(2t+1)^4-(2t-1)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}16t^4+32t^3+24t^2+8t+1-(16t^4-32t^3+24t^2-8t+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}16t^4+32t^3+24t^2+8t+1-16t^4+32t^3-24t^2+8t-1 \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{16t^4}+32t^3+ \cancel{24t^2}+8t+ \cancel{1} -\cancel{16t^4}+32t^3 -\cancel{24t^2}+8t -\cancel{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}64t^3+16t\end{aligned} $$ | |
| ① | $$ (2t+1)^4 = (2t+1)^2 \cdot (2t+1)^2 $$ |
| ② | Find $ \left(2t+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}{ 2t } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2t+1\right)^2 = \color{blue}{\left( 2t \right)^2} +2 \cdot 2t \cdot 1 + \color{red}{1^2} = 4t^2+4t+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{4t^2+4t+1}\right) $ by each term in $ \left( 4t^2+4t+1\right) $. $$ \left( \color{blue}{4t^2+4t+1}\right) \cdot \left( 4t^2+4t+1\right) = 16t^4+16t^3+4t^2+16t^3+16t^2+4t+4t^2+4t+1 $$ |
| ④ | Combine like terms: $$ 16t^4+ \color{blue}{16t^3} + \color{red}{4t^2} + \color{blue}{16t^3} + \color{green}{16t^2} + \color{orange}{4t} + \color{green}{4t^2} + \color{orange}{4t} +1 = \\ = 16t^4+ \color{blue}{32t^3} + \color{green}{24t^2} + \color{orange}{8t} +1 $$$$ (2t-1)^4 = (2t-1)^2 \cdot (2t-1)^2 $$ |
| ⑤ | Find $ \left(2t-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}{ 2t } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(2t-1\right)^2 = \color{blue}{\left( 2t \right)^2} -2 \cdot 2t \cdot 1 + \color{red}{1^2} = 4t^2-4t+1\end{aligned} $$ |
| ⑥ | Multiply each term of $ \left( \color{blue}{4t^2-4t+1}\right) $ by each term in $ \left( 4t^2-4t+1\right) $. $$ \left( \color{blue}{4t^2-4t+1}\right) \cdot \left( 4t^2-4t+1\right) = 16t^4-16t^3+4t^2-16t^3+16t^2-4t+4t^2-4t+1 $$ |
| ⑦ | Combine like terms: $$ 16t^4 \color{blue}{-16t^3} + \color{red}{4t^2} \color{blue}{-16t^3} + \color{green}{16t^2} \color{orange}{-4t} + \color{green}{4t^2} \color{orange}{-4t} +1 = \\ = 16t^4 \color{blue}{-32t^3} + \color{green}{24t^2} \color{orange}{-8t} +1 $$ |
| ⑧ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 16t^4-32t^3+24t^2-8t+1 \right) = -16t^4+32t^3-24t^2+8t-1 $$ |
| ⑨ | Combine like terms: $$ \, \color{blue}{ \cancel{16t^4}} \,+ \color{green}{32t^3} + \, \color{orange}{ \cancel{24t^2}} \,+ \color{red}{8t} + \, \color{green}{ \cancel{1}} \, \, \color{blue}{ -\cancel{16t^4}} \,+ \color{green}{32t^3} \, \color{orange}{ -\cancel{24t^2}} \,+ \color{red}{8t} \, \color{green}{ -\cancel{1}} \, = \color{green}{64t^3} + \color{red}{16t} $$ |
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