How do you simplify $(3 - 5 i) (4 + 6 i)$ $(3-5i)(4+6i)$ ?

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Answer 1 · 2018-08-08T01:37:49

$42 - 2 i$ $42 - 2i$

$(3 - 5 i) (4 + 6 i)$ $(3-5i)(4+6i)$

First multiply the first values:
$3 \cdot 4 = 12$ $3 * 4 = 12$

Outer values:
$3 \cdot 6 i = 18 i$ $3 * 6i = 18i$

Inner values:
$- 5 i \cdot 4 = - 20 i$ $-5i * 4 = -20i$

Last values:
$- 5 i \cdot 6 i = - 30 i^{2}$ $-5i * 6i = -30i^2$

Combine them all together:
$12 + 18 i - 20 i - 30 i^{2}$ $12 + 18i - 20i - 30i^2$

Combine like terms:
$12 - 2 i - 30 i^{2}$ $12 - 2i - 30i^2$

We know that $i^{2}$ $i^2$ is equivalent to $- 1$ $-1$ , so:
$12 - 2 i - 30 (- 1)$ $12 - 2i - 30(-1)$

Simplify:
$12 - 2 i + 30$ $12 - 2i + 30$

$42 - 2 i$ $42 - 2i$

Hope this helps!

How do you simplify (3-5i)(4+6i)(3−5i)(4+6i)(3-5i)(4+6i)?