Question

How do you find all the real and complex roots of `(x^3 + 9)(x^2 - 4) = 0`?

Answer 1

`x=-root3(9),+-2,root 3 9/2+-(3root6 3)/2i`

Explanation:

We can split this up into two separate equations:

`x^3+9=0" "" and "" "x^2-4=0`

Let's first solve the second equation (it's simpler):

`x^2-4=0`

`x^2=4`

`x=+-2`

These are the first `2` of what should be `5` total solutions.

Onto the next equation, we may be tempted to solve like so:

`x^3+9=0`

`x^3=-9`

`x=root3(-9)`

However, this doesn't deal with the two remaining zeros we should have. We must try to find a way to find the two leftover complex roots in `x^3+9=0`.

We can factor `x^3+9` as a sum of cubes, where the terms being cubed are `x` and `3^(2/3)`.

`x^3+9=0`

`(x+3^(2/3))(x^2-3^(2/3)x+3^(4/3))=0`

Solving the linear term gives us `x=-3^(2/3)=-root3 9`. To find the remaining two roots, use the quadratic formula on the quadratic factor.

`x=(3^(2/3)+-sqrt(3^(4/3)-4(1)(3^(4/3))))/2`

`x=(3^(2/3)+-sqrt(-3(3^(4/3))))/2`

`x=(3^(2/3)+-sqrt(3^(7/3))i)/2`

`x=3^(2/3)/2+-3^(7/6)/2i`

This can also be written as

`x=root 3 9/2+-(3root6 3)/2i`

or

`xapprox1.0400+-1.8014i`

depending on preference.