Question

How do you list all possible roots and find all factors of `x^3+27`?

Answer 1

`x^3+27 = (x+3)(x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)`

with corresponding zeros:

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

Explanation:

The sum of cubes identity can be written:

`a^3+b^3 = (a+b)(a^2-ab+b^2)`

We can use this with `a=x` and `b=3` as follows:

`x^3+27 = x^3+3^3`

`color(white)(x^3+27) = (x+3)(x^2-3x+9)`

So one linear factor is `(x+3)` with corresponding zero `x=-3`.

We can factor the remaining quadratic using Complex coeffients by completing the square and using the difference of squares identity, which can be written:

`a^2-b^2 = (a-b)(a+b)`

with `a=(x-3/2)` and `b = (3sqrt(3))/2 i` as follows:

`x^2-3x+9 = (x^2-3x+9/4)+27/4`

`color(white)(x^2-3x+9) = (x-3/2)^2-((3sqrt(3))/2 i)^2`

`color(white)(x^2-3x+9) = ((x-3/2)-(3sqrt(3))/2 i)((x-3/2)+(3sqrt(3))/2 i)`

`color(white)(x^2-3x+9) = (x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)`

Hence the remaining zeros are:

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

and the full factorisation can be written:

`x^3+27 = (x+3)(x-3/2-(3sqrt(3))/2 i)(x-3/2+(3sqrt(3))/2 i)`

Structly speaking, we have not listed all of the factors of `x^3+27`.

The ones we have identified so far are:

`x^3+27`

`x+3`

`x^2-3x+9`

`x-3/2-(3sqrt(3))/2i`

`x-3/2+(3sqrt(3))/2i`

`1`

In addition the results of multiplying these linear factors give us the other factors:

`(x+3)(x-3/2-(3sqrt(3))/2i) = x^2+(3/2-(3sqrt(3))/2i)x-(9/2+(9sqrt(3))/2i)`

`(x+3)(x-3/2+(3sqrt(3))/2i) = x^2+(3/2+(3sqrt(3))/2i)x-(9/2-(9sqrt(3))/2i)`

I don't think it was intended that you should list all of these factors, but there they are in case.