How do you write a polynomial function of least degree and leading coefficient 1 when the zeros are -2, -2, 3, -4i?

1 Answer
Jun 2, 2018

Answer:

Construct it from linear terms including the given roots

Explanation:

#x# minus each of the zeroes is equal to zero, so #f(x)=(x+2)^2(x-3)(x+4i)# does the job and has no more degrees than needed - assuming that the question-setter requires the repeated root at #x=-2#. If they don't, then #f(x)=(x+2)(x-3)(x+4i)# is better.

This multiplies out to be
#f(x)=x^3-(1-4i)x^2-(6+4i)x-24i#
in the non-repeated root case and
#f(x)=x^4+(1+ri)x^3-(8-4i)x^2-(12+32i)x-48i#
if we include the root at #x=-2# twice.