Question
Answer
$$(x-(3-i))(x-(3+i))=-i^2+x^2-6x+9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(x-(3-i))(x-(3+i))& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(x-3+i)(x-3-i) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-i^2+x^2-6x+9\end{aligned} $$ | |
| ① | Remove the parentheses by changing the sign of each term within them. $$ - \left( 3-i \right) = -3+i $$Remove the parentheses by changing the sign of each term within them. $$ - \left( 3+i \right) = -3-i $$ |
| ② | Multiply each term of $ \left( \color{blue}{x-3+i}\right) $ by each term in $ \left( x-3-i\right) $. $$ \left( \color{blue}{x-3+i}\right) \cdot \left( x-3-i\right) = \\ = x^2-3x -\cancel{ix}-3x+9+ \cancel{3i}+ \cancel{ix} -\cancel{3i}-i^2 $$ |
| ③ | Combine like terms: $$ x^2 \color{blue}{-3x} \, \color{red}{ -\cancel{ix}} \, \color{blue}{-3x} +9+ \, \color{orange}{ \cancel{3i}} \,+ \, \color{red}{ \cancel{ix}} \, \, \color{orange}{ -\cancel{3i}} \,-i^2 = -i^2+x^2 \color{blue}{-6x} +9 $$ |
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