How do you simplify $(-10+11i)(4+2i)-(4i)(-8i)(1-8i)$ ?

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Answer 1 · 2016-11-08T04:34:08

$-94+280i$

Simplify $(-10+11i)(4+2i)-(4i)(-8i)(1-8i)$

Let's start by multiplying $(-10+11i)(4+2i)$

Multiply each term in the first binomial by each term in the second.

$-40-20i+44i+22i^2 - (4i)(-8i)(1-8i)$

Combine like terms and recall that $i^2=-1$

$-40+24i+22(-1)-(4i)(-8i)(1-8i)$

$-40+24i-22-(4i)(-8i)(1-8i)$

$-62+24i-(4i)(-8i)(1-8i)$

Now let's multiply $(4i)(-8i)$

$-62+24i-(-32i^2)(1-8i)$

$-62+24i+32i^2(1-8i)$

Again, recall that $i^2=-1$

$-62+24i+32(-1)(1-8i)$

$-62+24i-32(1-8i)$

Distribute the $-32$

$-62+24i-32+256i$

Combine like terms

$-94+280i$

How do you simplify (-10+11i)(4+2i)-(4i)(-8i)(1-8i)(-10+11i)(4+2i)-(4i)(-8i)(1-8i)?