f(x)=(x+1)[x-(3-2i)]

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Solution for f(x)=(x+1)[x-(3-2i)] equation: 


Simplifying
f(x) = (x + 1)[x + -1(3 + -2i)]

Multiply f * x
fx = (x + 1)[x + -1(3 + -2i)]

Reorder the terms:
fx = (1 + x)[x + -1(3 + -2i)]
fx = (1 + x)[x + (3 * -1 + -2i * -1)]
fx = (1 + x)[x + (-3 + 2i)]

Reorder the terms:
fx = (1 + x)[-3 + 2i + x]

Multiply (1 + x) * [-3 + 2i + x]
fx = (1[-3 + 2i + x] + x[-3 + 2i + x])
fx = ([-3 * 1 + 2i * 1 + x * 1] + x[-3 + 2i + x])
fx = ([-3 + 2i + 1x] + x[-3 + 2i + x])
fx = (-3 + 2i + 1x + [-3 * x + 2i * x + x * x])

Reorder the terms:
fx = (-3 + 2i + 1x + [2ix + -3x + x2])
fx = (-3 + 2i + 1x + [2ix + -3x + x2])

Reorder the terms:
fx = (-3 + 2i + 2ix + 1x + -3x + x2)

Combine like terms: 1x + -3x = -2x
fx = (-3 + 2i + 2ix + -2x + x2)

Solving
fx = -3 + 2i + 2ix + -2x + x2

Solving for variable 'f'.

Move all terms containing f to the left, all other terms to the right.

Divide each side by 'x'.
f = -3x-1 + 2ix-1 + 2i + -2 + x

Simplifying
f = -3x-1 + 2ix-1 + 2i + -2 + x

Reorder the terms:
f = -2 + 2i + 2ix-1 + -3x-1 + x

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