Answer
$$(8+5i)\cdot(3+2i)-(4+i)\cdot(4-i)=31i-3$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(8+5i)\cdot(3+2i)-(4+i)\cdot(4-i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}24+16i+15i+10i^2-(16-4i+4i-i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}10i^2+31i+24-(-i^2+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-10+31i+24-(1+16) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}31i+14-17 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}31i-3\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{8+5i}\right) $ by each term in $ \left( 3+2i\right) $. $$ \left( \color{blue}{8+5i}\right) \cdot \left( 3+2i\right) = 24+16i+15i+10i^2 $$Multiply each term of $ \left( \color{blue}{4+i}\right) $ by each term in $ \left( 4-i\right) $. $$ \left( \color{blue}{4+i}\right) \cdot \left( 4-i\right) = 16 -\cancel{4i}+ \cancel{4i}-i^2 $$ |
| ② | Combine like terms: $$ 24+ \color{blue}{16i} + \color{blue}{15i} +10i^2 = 10i^2+ \color{blue}{31i} +24 $$Combine like terms: $$ 16 \, \color{blue}{ -\cancel{4i}} \,+ \, \color{blue}{ \cancel{4i}} \,-i^2 = -i^2+16 $$ |
| ③ | $$ 10i^2 = 10 \cdot (-1) = -10 $$$$ -i^2 = -(-1) = 1 $$ |
| ④ | Combine like terms: $$ \color{blue}{-10} +31i+ \color{blue}{24} = 31i+ \color{blue}{14} $$Combine like terms: $$ \color{blue}{1} + \color{blue}{16} = \color{blue}{17} $$ |
| ⑤ | Combine like terms: $$ 31i+ \color{blue}{14} \color{blue}{-17} = 31i \color{blue}{-3} $$ |
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