How do you multiply $(- 2 - 8 i) (6 + 7 i)$ $(-2-8i)(6+7i)$ ?

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How do you multiply $(- 2 - 8 i) (6 + 7 i)$ $(-2-8i)(6+7i)$ ?

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Answer 1 · 2015-02-10T00:47:14

You multiplly as in a normal multiplication remembering that $i = \sqrt{- 1}$ $i=sqrt(-1)$ and so $i^{2} = - 1$ $i^2=-1$ ;

so you get:

$(- 2 - 8 i) (6 + 7 i) =$ $(-2-8i)(6+7i)=$
$= - 12 - 14 i - 48 i - 56 i^{2} =$ $=-12-14i-48i-56i^2=$
$= - 12 - 62 i - (- 1 \cdot 56) = 44 - 62 i$ $=-12-62i-(-1*56)=44-62i$

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How do you multiply $(- 2 - 8 i) (6 + 7 i)$ $(-2-8i)(6+7i)$ ?

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How do you multiply (−2−8i)(6+7i)(-2-8i)(6+7i)?

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How do you multiply $(- 2 - 8 i) (6 + 7 i)$ $(-2-8i)(6+7i)$ ?