How do you find all zeros of the function #f(x)= x^4 + 5x^2 + 4# given -i as a zero?

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Answer 1 · 2016-04-06T23:19:12

Factor by grouping to find:

#x = +-i# or #x = +-2i#

This polynomial factors by grouping as follows:

#f(x) = x^4+5x^2+4#

#= x^4+4x^2+x^2+4#

#= x^2(x^2+4)+1(x^2+4)#

#= (x^2+1)(x^2+4)#

#= (x^2-i^2)(x^2-(2i)^2)#

#= (x-i)(x+i)(x-2i)(x+2i)#

Hence zeros: #x = +-i# and #x = +-2i#

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Another way of finding this is to note that since #f(x)# has Real coefficients, any zeros must occur in Complex conjugate pairs.

So if #x = -i# is a zero then so is #x = i#

Therefore #f(x)# is divisible by #(x+i)(x-i) = x^2-i^2 = x^2+1#

Then dividing #f(x)# by #x^2+1# we find:

#f(x) = (x^2+1)(x^2+4)#

Hence the other two zeros are #+-2i#