Question

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

Answer 1

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`