(x-3)(x-2i)(x-x+2i)=0

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Solution for (x-3)(x-2i)(x-x+2i)=0 equation:


Simplifying
(x + -3)(x + -2i)(x + -1x + 2i) = 0

Reorder the terms:
(-3 + x)(x + -2i)(x + -1x + 2i) = 0

Reorder the terms:
(-3 + x)(-2i + x)(x + -1x + 2i) = 0

Reorder the terms:
(-3 + x)(-2i + x)(2i + x + -1x) = 0

Combine like terms: x + -1x = 0
(-3 + x)(-2i + x)(2i + 0) = 0
(-3 + x)(-2i + x)(2i) = 0

Remove parenthesis around (2i)
(-3 + x)(-2i + x) * 2i = 0

Reorder the terms for easier multiplication:
2i(-3 + x)(-2i + x) = 0

Multiply (-3 + x) * (-2i + x)
2i(-3(-2i + x) + x(-2i + x)) = 0
2i((-2i * -3 + x * -3) + x(-2i + x)) = 0
2i((6i + -3x) + x(-2i + x)) = 0
2i(6i + -3x + (-2i * x + x * x)) = 0
2i(6i + -3x + (-2ix + x2)) = 0

Reorder the terms:
2i(6i + -2ix + -3x + x2) = 0
2i(6i + -2ix + -3x + x2) = 0
(6i * 2i + -2ix * 2i + -3x * 2i + x2 * 2i) = 0

Reorder the terms:
(-6ix + 2ix2 + 12i2 + -4i2x) = 0
(-6ix + 2ix2 + 12i2 + -4i2x) = 0

Solving
-6ix + 2ix2 + 12i2 + -4i2x = 0

Solving for variable 'i'.

Factor out the Greatest Common Factor (GCF), '2i'.
2i(-3x + x2 + 6i + -2ix) = 0

Ignore the factor 2.

Subproblem 1

Set the factor 'i' equal to zero and attempt to solve: Simplifying i = 0 Solving i = 0 Move all terms containing i to the left, all other terms to the right. Simplifying i = 0

Subproblem 2

Set the factor '(-3x + x2 + 6i + -2ix)' equal to zero and attempt to solve: Simplifying -3x + x2 + 6i + -2ix = 0 Reorder the terms: 6i + -2ix + -3x + x2 = 0 Solving 6i + -2ix + -3x + x2 = 0 Move all terms containing i to the left, all other terms to the right. Add '3x' to each side of the equation. 6i + -2ix + -3x + 3x + x2 = 0 + 3x Combine like terms: -3x + 3x = 0 6i + -2ix + 0 + x2 = 0 + 3x 6i + -2ix + x2 = 0 + 3x Remove the zero: 6i + -2ix + x2 = 3x Add '-1x2' to each side of the equation. 6i + -2ix + x2 + -1x2 = 3x + -1x2 Combine like terms: x2 + -1x2 = 0 6i + -2ix + 0 = 3x + -1x2 6i + -2ix = 3x + -1x2 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined.

Solution

i = {0}

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