Question
Answer
$$(6+xi)\cdot(5+2i)+(6+xi)\cdot(3-2i)=8ix+48$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6+xi)\cdot(5+2i)+(6+xi)\cdot(3-2i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}30+12i+5ix+2i^2x+18-12i+3ix-2i^2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}8ix+48\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{6+ix}\right) $ by each term in $ \left( 5+2i\right) $. $$ \left( \color{blue}{6+ix}\right) \cdot \left( 5+2i\right) = 30+12i+5ix+2i^2x $$Multiply each term of $ \left( \color{blue}{6+ix}\right) $ by each term in $ \left( 3-2i\right) $. $$ \left( \color{blue}{6+ix}\right) \cdot \left( 3-2i\right) = 18-12i+3ix-2i^2x $$ |
| ② | Combine like terms: $$ \color{blue}{30} + \, \color{red}{ \cancel{12i}} \,+ \color{orange}{5ix} + \, \color{blue}{ \cancel{2i^2x}} \,+ \color{blue}{18} \, \color{red}{ -\cancel{12i}} \,+ \color{orange}{3ix} \, \color{blue}{ -\cancel{2i^2x}} \, = \color{orange}{8ix} + \color{blue}{48} $$ |
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