How do you write a polynomial with zeros 2, -2, -6i and leading coefficient 1?

1 Answer
May 10, 2017

Answer:

Identify the four (not three) zeroes and write out the zeroes in factored form. Answer: #x^4+32x^2-144=0#

Explanation:

Original question: Find polynomial with zeroes #2, -2, -6i# and leading coefficient of #1#

We know that each complex zero must have a pair that is also a complex zero, which is its conjugate. Therefore, if #-6i# is a zero, then #6i# is a zero too.

We also know that zeroes of a polynomial come in the form #(x-a)(x-b)(x-c)(x-d)=0# where #a,b,c,d# are zeroes, therefore we can write our polynomial in this same form:
#(x-2)(x+2)(x+6i)(x-6i)=0#

Now, we can write out the expanded form by using the fact that #(a+b)(a-b)=a^2-b^2# and expand the resulting product of binomials:
Note that #(6i)^2=36(-1)=-36#
#(x^2-4)(x^2+36)=0#
#x^2*x^2+x^2*36-4*x^2-4*36=0#
#x^4+36x^2-4x^2-144=0#

Combining like-terms, we get:
#x^4+32x^2-144=0#