How do you simplify $(-3-i)(2-2i)$ ?

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Answer 1 · 2016-10-20T14:22:52

$-8+4i$

Multiply the brackets out as for normal algebra but replace $i^2$ with $-1$

$(-3-i)(2-2i)$

multiplying out, and building up the expansion.

$(color(red)(-3)-i)(color(red)2-2i)$

= $color(red)(-6)$

$(color(red)(-3)-i)(2-color(red)(2i))$

= $-6-(color(red)(-6i))$

$(-3-color(red)(i))(color(red)2-2i)$

= $-6-(-6i)-color(red)(2i)$

$(-3-color(red)(i))(2-color(red)(2i))$

= $-6-(-6i)-2i+color(red)(2i^2)$

$=-6+6i-2i+2xx(-1)$

$=-6+6i-2i-2$

collecting like terms, putting the real part first , we have:

$-8+4i$

How do you simplify (-3-i)(2-2i)(-3-i)(2-2i)?