Question
Answer
$$(1+3xi)\cdot(3-i)-(5+3xi)\cdot(3-i)=4i-12$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+3xi)\cdot(3-i)-(5+3xi)\cdot(3-i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}3-i+9ix-3i^2x-(15-5i+9ix-3i^2x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3-i+9ix-3i^2x-15+5i-9ix+3i^2x \xlongequal{ } \\[1 em] & \xlongequal{ }3-i+ \cancel{9ix} -\cancel{3i^2x}-15+5i -\cancel{9ix}+ \cancel{3i^2x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4i-12\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+3ix}\right) $ by each term in $ \left( 3-i\right) $. $$ \left( \color{blue}{1+3ix}\right) \cdot \left( 3-i\right) = 3-i+9ix-3i^2x $$Multiply each term of $ \left( \color{blue}{5+3ix}\right) $ by each term in $ \left( 3-i\right) $. $$ \left( \color{blue}{5+3ix}\right) \cdot \left( 3-i\right) = 15-5i+9ix-3i^2x $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 15-5i+9ix-3i^2x \right) = -15+5i-9ix+3i^2x $$ |
| ③ | Combine like terms: $$ \color{blue}{3} \color{red}{-i} + \, \color{green}{ \cancel{9ix}} \, \, \color{blue}{ -\cancel{3i^2x}} \, \color{blue}{-15} + \color{red}{5i} \, \color{green}{ -\cancel{9ix}} \,+ \, \color{blue}{ \cancel{3i^2x}} \, = \color{red}{4i} \color{blue}{-12} $$ |
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