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

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Answer 1 · 2016-10-20T14:22:52

$- 8 + 4 i$ $-8+4i$

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

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

multiplying out, and building up the expansion.

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

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

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

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

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

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

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

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

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

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

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

$- 8 + 4 i$ $-8+4i$

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