Question

Write the equation in standard form of the polynomial that has zeros at x=±6i

Answer 1

`f(x)=x^2+36`

Explanation:

We want to find the equation in standard form for a polynomial that has zeros at x=6i and x=-6i.

So, we will have the two factors:

`(x-(6i))\text{ and } (x-(-6i))`

So, our polynomial will be:

`f(x)=(x-6i)(x+6i)`

Distribute:

`f(x)=x(x+6i)-6i(x+6i)`

Distribute:

`f(x)=x^2+6xi-6xi-36i^2`

Combine like terms:

`f(x)=x^2-36i^2`

Remember that i²=-1. Hence:

`f(x)=x^2-(-36)`

Simplify. So, our polynomial is:

`f(x)=x^2+36`