Question
Answer
$$(1+4xi)\cdot(4+i)-(1+4xi)\cdot(4-3i)=16i^2x+4i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+4xi)\cdot(4+i)-(1+4xi)\cdot(4-3i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}4+i+16ix+4i^2x-(4-3i+16ix-12i^2x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4+i+16ix+4i^2x-4+3i-16ix+12i^2x \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{4}+i+ \cancel{16ix}+4i^2x -\cancel{4}+3i -\cancel{16ix}+12i^2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}16i^2x+4i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+4ix}\right) $ by each term in $ \left( 4+i\right) $. $$ \left( \color{blue}{1+4ix}\right) \cdot \left( 4+i\right) = 4+i+16ix+4i^2x $$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 | Remove the parentheses by changing the sign of each term within them. $$ - \left( 4-3i+16ix-12i^2x \right) = -4+3i-16ix+12i^2x $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{4}} \,+ \color{green}{i} + \, \color{orange}{ \cancel{16ix}} \,+ \color{red}{4i^2x} \, \color{blue}{ -\cancel{4}} \,+ \color{green}{3i} \, \color{orange}{ -\cancel{16ix}} \,+ \color{red}{12i^2x} = \color{red}{16i^2x} + \color{green}{4i} $$ |
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