How do you evaluate $- 2 i (- 2 + 3 i)$ $-2i ( - 2+ 3i )$ ?

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Answer 1 · 2017-08-02T17:49:25

$6 + 4 i$ $6+4i$

To evaluate this, distribute $- 2 i$ $-2i$ through the expression $- 2 + 3 i$ $-2+3i$ .

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

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

$4 i + (- 6 i^{2})$ $4i+(-6i^2)$

$4 i - 6 i^{2}$ $4i-6i^2$

Since we know that $i = \sqrt{- 1}$ $i=sqrt(-1)$ , $i^{2} = {(\sqrt{- 1})}^{2} = - 1$ $i^2=(sqrt(-1))^2 = -1$ .

$4 i - 6 (- 1)$ $4i-6(-1)$

$4 i + 6$ $4i+6$

Rewriting this in $a + b i$ $a+bi$ form, we get $6 + 4 i$ $6+4i$ as the answer.

How do you evaluate −2i(−2+3i)-2i ( - 2+ 3i )?