How do you write the polynomial function given the following zeros: -5i, -i, i, 2i?

2 Answers

It is #P(x)=(x+5i)(x+i)(x-i)(x-2i)=x^4+3ix^3+11x^2+3ix+10#

where #i# is the imaginery unit (#i^2=-1#)

Sep 17, 2015

Answer:

If the polynomial also has Real coefficients, then it also has zeroes #5i# and #-2i#

The simplest such polynomial is:

#f(x) = (x-5i)(x+5i)(x-i)(x+i)(x-2i)(x+2i)#

#=x^6+30x^4+129x^2+100#

Explanation:

The Complex zeroes of polynomials with Real coefficients always occur in conjugate pairs.

So in our example, since #-5i# and #2i# are zeroes, so are #5i# and #-2i#.

Hence:

#f(x) = (x-5i)(x+5i)(x-i)(x+i)(x-2i)(x+2i)#

#= (x^2+25)(x^2+1)(x^2+4)#

#=x^6+(25+1+4)x^4+(1*4+25*4+25*1)x^2+(25*1*4)#

#=x^6+30x^4+129x^2+100#

Any polynomial in #x# with these roots will be a multiple (scalar or polynomial) of this #f(x)#.