Question
Answer
$$(-x)(-2+4i)^2=16ix+12x$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-x)(-2+4i)^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(-x)(4-16i+16i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(-x)(4-16i-16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(-x)(-16i-12) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}16ix+12x\end{aligned} $$ | |
| ① | Find $ \left(-2+4i\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply formula $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 2 } $ and $ B = \color{red}{ 4i }$. $$ \begin{aligned}\left(-2+4i\right)^2& \xlongequal{ S1 } \left(2-4i\right)^2 \xlongequal{ S2 } \color{blue}{2^2} -2 \cdot 2 \cdot 4i + \color{red}{\left( 4i \right)^2} = \\[1 em] & = 4-16i+16i^2\end{aligned} $$ |
| ② | $$ 16i^2 = 16 \cdot (-1) = -16 $$ |
| ③ | Combine like terms: $$ \color{blue}{4} -16i \color{blue}{-16} = -16i \color{blue}{-12} $$ |
| ④ | Multiply $ \color{blue}{-x} $ by $ \left( -16i-12\right) $ $$ \color{blue}{-x} \cdot \left( -16i-12\right) = 16ix+12x $$ |
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