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Conjugate Pair Theorem
----------------------

If P is a polynomial in one variable with real coefficients,
and a + bi is a root of P with a and b real numbers, then
its complex conjugate a - bi  is also a root of P.

Example:

P(x) = x3 - 9x2 + 16x - 14 = 0

With x = 1 + i

∴ (x - (1 + i))(x - (1 - i))(x ± ?) = x3 - 9x2 + 16x - 14

∴ (x - 1 - i)(x - 1 + i)g(x) = x3 - 9x2 + 16x - 14

∴ (x2 - x + ix - x + 1 - i - ix + i + 1)g(x) = x3 - 9x2 + 16x - 14

∴ (x2 - 2x + 2)g(x) = x3 - 9x2 + 16x - 14

∴ g(x) = (x3 - 9x2 + 16x - 14)/(x2 - 2x + 2)

∴ g(x) = (x - 7) after long division.

The factors are:

(x - (1 + i))(x - (1 - i))(x - 7)