How do you simplify $(2 - 2 i) \cdot (- 3 + 3 i)$ $(2-2i)*(-3+3i)$ ?

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Answer 1 · 2017-01-04T13:10:58

Perform the multiplication, using the F.O.I.L. method, substitute -1 for $i^{2}$ $i^2$ , and then combine like terms.

Multiply the F irst terms:

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6$ $(2 -2i)*(-3 + 3i) = -6$

Multiply the Outside terms:

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6 + 6 i$ $(2 -2i)*(-3 + 3i) = -6 + 6i$

Multiply the I nside terms:

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6 + 6 i + 6 i$ $(2 -2i)*(-3 + 3i) = -6 + 6i + 6i$

Multiply the Last terms:

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6 + 6 i + 6 i - 9 i^{2}$ $(2 -2i)*(-3 + 3i) = -6 + 6i + 6i - 9i^2$

Substitute -1 for $i^{2}$ $i^2$ :

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6 + 6 i + 6 i - 9 (- 1)$ $(2 -2i)*(-3 + 3i) = -6 + 6i + 6i - 9(-1)$

$(2 - 2 i) \cdot (- 3 + 3 i) = - 6 + 6 i + 6 i + 9$ $(2 -2i)*(-3 + 3i) = -6 + 6i + 6i + 9$

Combine like terms:

$(2 - 2 i) \cdot (- 3 + 3 i) = 3 + 12 i$ $(2 -2i)*(-3 + 3i) = 3 + 12i$

I hope that this helps.

How do you simplify (2-2i)*(-3+3i)(2−2i)⋅(−3+3i)(2-2i)*(-3+3i)?