How do you multiply $3 i (4 + 2 i) (2 + 5 i)$ $3i(4+2i)(2+5i)$ ?

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Answer 1 · 2017-02-09T16:13:03

The answer is $= - 72 - 6 i$ $=-72-6i$

Remember that

$i^{2} = - 1$ $i^2=-1$

Therefore,

$(4 + 2 i) (2 + 5 i) = 8 + 20 i + 4 i + 10 i^{2}$ $(4+2i)(2+5i)=8+20i+4i+10i^2$

$= 8 - 10 + 24 i$ $=8-10+24i$

$= - 2 + 24 i$ $=-2+24i$

so,

$3 i (4 + 2 i) (2 + 5 i) = 3 i (- 2 + 24 i)$ $3i(4+2i)(2+5i)=3i(-2+24i)$

$= - 6 i + 72 i^{2}$ $=-6i+72i^2$

$= - 72 - 6 i$ $=-72-6i$

How do you multiply 3i(4+2i)(2+5i)3i(4+2i)(2+5i)3i(4+2i)(2+5i)?