# How do you perform the operation and write the result in standard form given (6-2i)(2-3i)?

Nov 27, 2016

$6 - 22 i$

#### Explanation:

The standard form is $\textcolor{b l u e}{a} + \textcolor{red}{b} i$

$\left(6 - 2 i\right) \left(2 - 3 i\right) = 12 - 18 i - 4 i + 6 {i}^{2}$

$= 12 - 22 i + 6 {i}^{2}$
Apply the definition ${i}^{2} = - 1$
$= 12 - 22 i + 6 \left(- 1\right)$
$= 12 - 22 i - 6$
$= \textcolor{b l u e}{6} \textcolor{red}{-} \textcolor{red}{22} i$