How do you simplify $4 i (- 2 - 8 i)$ $4i(-2-8i)$ ?

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How do you simplify $4 i (- 2 - 8 i)$ $4i(-2-8i)$ ?

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Answer 1 · 2018-04-15T02:25:31

The answer is $- 8 i + 32$ $-8i+32$ .

Explanation:

Simplifying $4 i (- 2 - 8 i)$ $4i(-2-8i)$ :

First, factor out the $4 i$ $4i$ : $- 8 i - 32 i^{2}$ $-8i-32i^2$ .

Because $i$ $i$ , an imaginary number, is equal to $\sqrt{- 1}$ $sqrt(-1)$ , $i^{2}$ $i^2$ (or ${(\sqrt{- 1})}^{2}$ $(sqrt(-1))^2$ should be $- 1$ $-1$ .

Therefore, you can substitute in $- 1$ $-1$ for $i^{2}$ $i^2$ . Therefore your answer is $- 8 i - (32 \cdot - 1)$ $-8i-(32*-1)$ . This simplifies to your final answer, $- 8 i + 32$ $-8i+32$ .

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How do you simplify $4 i (- 2 - 8 i)$ $4i(-2-8i)$ ?

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