How do you find all roots #x^4 - 4x^3 + 6x^2 - 4x + 5# given one root 2 -i?

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Answer 1 · 2016-08-14T13:23:27

This quartic has zeros #2+-i# and #+-i#

#f(x) = x^4-4x^3+6x^2-4x+5#

Since the coefficients of #f(x)# are all Real, any Complex zeros must occur in Complex conjugate pairs.

So if #2-i# is a zero then so is #2+i#, and we have corresponding factors #(x-2+i)# and #(x-2-i)#

Then:

#(x-2+i)(x-2-i) = (x-2)^2-i^2 = x^2-4x+5#

We find:

#x^4-4x^3+6x^2-4x+5 = (x^2-4x+5)(x^2+1)#

The remaining zeros, which are those of #x^2+1# are simply #+-i#