Question
Answer
$$(3y^2+1)^4=81y^8+108y^6+54y^4+12y^2+1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3y^2+1)^4& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}81y^8+108y^6+54y^4+12y^2+1\end{aligned} $$ | |
| ① | $$ (3y^2+1)^4 = (3y^2+1)^2 \cdot (3y^2+1)^2 $$ |
| ② | Find $ \left(3y^2+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}{ 3y^2 } $ and $ B = \color{red}{ 1 }$. $$ \begin{aligned}\left(3y^2+1\right)^2 = \color{blue}{\left( 3y^2 \right)^2} +2 \cdot 3y^2 \cdot 1 + \color{red}{1^2} = 9y^4+6y^2+1\end{aligned} $$ |
| ③ | Multiply each term of $ \left( \color{blue}{9y^4+6y^2+1}\right) $ by each term in $ \left( 9y^4+6y^2+1\right) $. $$ \left( \color{blue}{9y^4+6y^2+1}\right) \cdot \left( 9y^4+6y^2+1\right) = 81y^8+54y^6+9y^4+54y^6+36y^4+6y^2+9y^4+6y^2+1 $$ |
| ④ | Combine like terms: $$ 81y^8+ \color{blue}{54y^6} + \color{red}{9y^4} + \color{blue}{54y^6} + \color{green}{36y^4} + \color{orange}{6y^2} + \color{green}{9y^4} + \color{orange}{6y^2} +1 = \\ = 81y^8+ \color{blue}{108y^6} + \color{green}{54y^4} + \color{orange}{12y^2} +1 $$ |
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