Question
Answer
$$(12-a+i\cdot(9+b))^2=b^2i^2-2abi+18bi^2+a^2-18ai+24bi+81i^2-24a+216i+144$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(12-a+i\cdot(9+b))^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(12-a+9i+bi)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1bi-a+9i+12)^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}b^2i^2-2abi+18bi^2+a^2-18ai+24bi+81i^2-24a+216i+144\end{aligned} $$ | |
| ① | Multiply $ \color{blue}{i} $ by $ \left( 9+b\right) $ $$ \color{blue}{i} \cdot \left( 9+b\right) = 9i+bi $$ |
| ② | Combine like terms: $$ 12-a+9i+bi = bi-a+9i+12 $$ |
| ③ | Multiply each term of $ \left( \color{blue}{bi-a+9i+12}\right) $ by each term in $ \left( bi-a+9i+12\right) $. $$ \left( \color{blue}{bi-a+9i+12}\right) \cdot \left( bi-a+9i+12\right) = \\ = b^2i^2-abi+9bi^2+12bi-abi+a^2-9ai-12a+9bi^2-9ai+81i^2+108i+12bi-12a+108i+144 $$ |
| ④ | Combine like terms: $$ b^2i^2 \color{blue}{-abi} + \color{red}{9bi^2} + \color{green}{12bi} \color{blue}{-abi} +a^2 \color{orange}{-9ai} \color{blue}{-12a} + \color{red}{9bi^2} \color{orange}{-9ai} +81i^2+ \color{red}{108i} + \color{green}{12bi} \color{blue}{-12a} + \color{red}{108i} +144 = \\ = b^2i^2 \color{blue}{-2abi} + \color{red}{18bi^2} +a^2 \color{orange}{-18ai} + \color{green}{24bi} +81i^2 \color{blue}{-24a} + \color{red}{216i} +144 $$ |
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