How do you multiply $-2i(3-7i)(4+2i)$ ?

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Answer 1 · 2016-09-04T16:50:31

$(-2i)(3-7i)(4+2i)=-44-52i$

While multiplying complex numbers we should always remember that $i^2=-1$ .

Hence, $(-2i)(3-7i)(4+2i)$

= $((-2i×3)-2i×(-7i))(4+2i)$

= $(-6i+14i^2)(4+2i)$

= $(-6i+14×(-1))(4+2i)$

= $(-14-6i)(4+2i)$

= $-14×4-14×2i-6i×4-6i×2i$

= $-56-28i-24i-12×(-1)$

= $-56-28i-24i+12$

= $-44-52i$

How do you multiply -2i(3-7i)(4+2i)-2i(3-7i)(4+2i)?