Question

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

Answer 1

Answer:

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

Step-by-step 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:

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