Question
Answer
$$-4n^4+6n^3+12abn^2-12abn+2n^2-(nb+n+ar)^2=-a^2r^2+12abn^2-2abnr-b^2n^2-4n^4-12abn-2anr-2bn^2+6n^3+n^2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-4n^4+6n^3+12abn^2-12abn+2n^2-(nb+n+ar)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-4n^4+6n^3+12abn^2-12abn+2n^2-(1a^2r^2+2abnr+b^2n^2+2anr+2bn^2+n^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-4n^4+6n^3+12abn^2-12abn+2n^2-a^2r^2-2abnr-b^2n^2-2anr-2bn^2-n^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-a^2r^2+12abn^2-2abnr-b^2n^2-4n^4-12abn-2anr-2bn^2+6n^3+n^2\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{bn+n+ar}\right) $ by each term in $ \left( bn+n+ar\right) $. $$ \left( \color{blue}{bn+n+ar}\right) \cdot \left( bn+n+ar\right) = b^2n^2+bn^2+abnr+bn^2+n^2+anr+abnr+anr+a^2r^2 $$ |
| ② | Combine like terms: $$ b^2n^2+ \color{blue}{bn^2} + \color{red}{abnr} + \color{blue}{bn^2} +n^2+ \color{green}{anr} + \color{red}{abnr} + \color{green}{anr} +a^2r^2 = \\ = a^2r^2+ \color{red}{2abnr} +b^2n^2+ \color{green}{2anr} + \color{blue}{2bn^2} +n^2 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( a^2r^2+2abnr+b^2n^2+2anr+2bn^2+n^2 \right) = -a^2r^2-2abnr-b^2n^2-2anr-2bn^2-n^2 $$ |
| ④ | Combine like terms: $$ -4n^4+6n^3+12abn^2-12abn+ \color{blue}{2n^2} -a^2r^2-2abnr-b^2n^2-2anr-2bn^2 \color{blue}{-n^2} = \\ = -a^2r^2+12abn^2-2abnr-b^2n^2-4n^4-12abn-2anr-2bn^2+6n^3+ \color{blue}{n^2} $$ |
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