Answer
$$(3+4xi)\cdot(4-3i)+(1-4xi)\cdot(4-3i)=-12i+16$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(3+4xi)\cdot(4-3i)+(1-4xi)\cdot(4-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}12-9i+16ix-12i^2x+4-3i-16ix+12i^2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-12i+16\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{3+4ix}\right) $ by each term in $ \left( 4-3i\right) $. $$ \left( \color{blue}{3+4ix}\right) \cdot \left( 4-3i\right) = 12-9i+16ix-12i^2x $$Multiply each term of $ \left( \color{blue}{1-4ix}\right) $ by each term in $ \left( 4-3i\right) $. $$ \left( \color{blue}{1-4ix}\right) \cdot \left( 4-3i\right) = 4-3i-16ix+12i^2x $$ |
| ② | Combine like terms: $$ \color{blue}{12} \color{red}{-9i} + \, \color{green}{ \cancel{16ix}} \, \, \color{blue}{ -\cancel{12i^2x}} \,+ \color{blue}{4} \color{red}{-3i} \, \color{green}{ -\cancel{16ix}} \,+ \, \color{blue}{ \cancel{12i^2x}} \, = \color{red}{-12i} + \color{blue}{16} $$ |
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