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