# How do you multiply (2i)(1-4i)(1+i)?

Mar 12, 2018

$\left(6 + 10 i\right)$

#### Explanation:

The rules of polynomial (brackets) multiplication are required, taking care to respect the rules of multiplying real and imaginary numbers. There are three brackets and you have to start with two of them but it does not matter which two as multiplication of complex numbers satisfies the distributive law.

Taking the first two brackets first (and showing many steps just to emphasise the point at this stage)

$\left(2 i\right) \left(1 - 4 i\right) \left(1 + i\right)$

$= \left(2 i \times 1 - 2 i \times 4 i\right) \left(1 + i\right)$

$= \left(2 i - \left(2 \times 4\right) \left(i \times i\right)\right) \left(1 + i\right)$

$= \left(2 i - \left(8\right) \left(- 1\right)\right) \left(1 + i\right)$

$= \left(2 i + 8\right) \left(1 + i\right)$

$= \left(8 + 2 i\right) \left(1 + i\right)$
(reordering the real and imaginary parts to the conventional form)

This leaves two remaining brackets to evaluate (now showing fewer steps)

$= \left(8 \times 1 + 8 \times i + 2 i \times 1 + 2 i \times i\right)$

$= \left(8 + 8 i + 2 i - 2\right)$

$= \left(6 + 10 i\right)$