Question
Answer
$$2-3i+(x-4i)^2-(3i+4x^2)=16i^2-8ix-3x^2-6i+2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}2-3i+(x-4i)^2-(3i+4x^2)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}2-3i+x^2-8ix+16i^2-(3i+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}16i^2-8ix+x^2-3i+2-(3i+4x^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}16i^2-8ix+x^2-3i+2-3i-4x^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16i^2-8ix-3x^2-6i+2\end{aligned} $$ | |
| ① | Find $ \left(x-4i\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ x } $ and $ B = \color{red}{ 4i }$. $$ \begin{aligned}\left(x-4i\right)^2 = \color{blue}{x^2} -2 \cdot x \cdot 4i + \color{red}{\left( 4i \right)^2} = x^2-8ix+16i^2\end{aligned} $$ |
| ② | Combine like terms: $$ 2-3i+x^2-8ix+16i^2 = 16i^2-8ix+x^2-3i+2 $$ |
| ③ | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3i+4x^2 \right) = -3i-4x^2 $$ |
| ④ | Combine like terms: $$ 16i^2-8ix+ \color{blue}{x^2} \color{red}{-3i} +2 \color{red}{-3i} \color{blue}{-4x^2} = 16i^2-8ix \color{blue}{-3x^2} \color{red}{-6i} +2 $$ |
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