how do i write the cubic equation that has roots of 8 and 2i in standard form

Question

phi · Answer

cubic means it will have x^3 in it.  It will have 3 roots.  it will have 2 complex roots.  The complex roots will be complex conjugates.
complex conjugate means if you have:   a+bi (a and b are numbers, i=sqrt(-1)), then the complex conjugate is a-bi

so your roots for a cubic will be:  8, 2i, -2i
that means your cubic is formed by (x-8)(x-2i)(x- -2i)=
(x-8)(x-2i)(x+2i)

now the hard part, multiply this out.

precal · Answer

(x-2i)(x+2i) is a difference of squares
$$x^2-\left( 2i \right)^2$$
$$x^2-4i^2$$
$$x^2-(4)(-1)$$
$$x^2+4$$
$$(x-8)(x^2+4)   $$
finish multiplying it out

phi · Answer

I left out the obvious :
(x-8)(x-2i)(x+2i)=0

we want an equation when we are done

precal · Answer

Yes I know, I just thought I would help by multiplying out (x+2i)(x-2i).
If they take my last step and multiply it out they will have their cubic equation.
It is always good to do some work.....