In order to have a zero of -5i, you also need one of 5i.
(x - 1)(x - 1)(x + 5i)(x - 5i) = (x2 -- 2x + 1)(x2 + 25i2) = x4 - 25x2 - 2x3 + 50x + x2 - 25 = x4 - 2x3 + 24x2 + 50x - 25
Lindsay D.
asked 10/18/20In order to have a zero of -5i, you also need one of 5i.
(x - 1)(x - 1)(x + 5i)(x - 5i) = (x2 -- 2x + 1)(x2 + 25i2) = x4 - 25x2 - 2x3 + 50x + x2 - 25 = x4 - 2x3 + 24x2 + 50x - 25
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