Question
Answer
$$(4-xi)\cdot(2+3i)-(4-xi)\cdot(2-5i)=-8i^2x+32i$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(4-xi)\cdot(2+3i)-(4-xi)\cdot(2-5i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}8+12i-2ix-3i^2x-(8-20i-2ix+5i^2x) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}8+12i-2ix-3i^2x-8+20i+2ix-5i^2x \xlongequal{ } \\[1 em] & \xlongequal{ } \cancel{8}+12i -\cancel{2ix}-3i^2x -\cancel{8}+20i+ \cancel{2ix}-5i^2x \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-8i^2x+32i\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{4-ix}\right) $ by each term in $ \left( 2+3i\right) $. $$ \left( \color{blue}{4-ix}\right) \cdot \left( 2+3i\right) = 8+12i-2ix-3i^2x $$Multiply each term of $ \left( \color{blue}{4-ix}\right) $ by each term in $ \left( 2-5i\right) $. $$ \left( \color{blue}{4-ix}\right) \cdot \left( 2-5i\right) = 8-20i-2ix+5i^2x $$ |
| ② | Remove the parentheses by changing the sign of each term within them. $$ - \left( 8-20i-2ix+5i^2x \right) = -8+20i+2ix-5i^2x $$ |
| ③ | Combine like terms: $$ \, \color{blue}{ \cancel{8}} \,+ \color{green}{12i} \, \color{orange}{ -\cancel{2ix}} \, \color{red}{-3i^2x} \, \color{blue}{ -\cancel{8}} \,+ \color{green}{20i} + \, \color{orange}{ \cancel{2ix}} \, \color{red}{-5i^2x} = \color{red}{-8i^2x} + \color{green}{32i} $$ |
This page was created using
Complex Expression calculator