Question

How do you factor completely `p(x)=(x^3)-(6x^2)-4x-10`?

Answer 1

`p(x) = (x-x_1)(x-x_2)(x-x_3)`

where `x_1`, `x_2` and `x_3` are as below:

Explanation:

`p(x) = x^3-6x^2-4x-10`

Attempt to solve `p(x) = 0`

Let `t = x-2`

Then:

`t^3 - 16t -34 = (x-2)^3 -16(x-2)-34`

`= x^3-6x^2+12x-8-16x+32-34`

`= x^3-6x^2-4x-10 = p(x) = 0`

Let `t = u+v`

`0 = (u+v)^3-16(u+v)-34`

`= u^3+v^3+(3uv-16)(u+v)-34`

Add the constraint `v = 16/(3u)` to make `(3uv-16) = 0`

Then:

`0 = u^3+(16/(3u))^3-34 = u^3+4096/(27u^3)-34`

Multiply both ends by `27u^3` to get:

`0 = 27(u^3)^2 - 918(u^3) + 4096`

Use the quadratic formula to get:

`u^3 = (918+-sqrt(918^2-(4*27*4096)))/(2*27)`

`= (918+-sqrt(400396))/54`

`= (918+-6sqrt(11121))/54`

`= (3(153+-sqrt(11121)))/27`

The derivation is symmetric in `u` and `v`, hence:

`t = root(3)((3(153+sqrt(11121)))/27) + root(3)((3(153-sqrt(11121)))/27)`

`= 1/3(root(3)(3(153+sqrt(11121))) + root(3)(3(153-sqrt(11121))))`

So the Real root of `p(x) = 0` is:

`x_1 = 2+t`

`= 2+1/3(root(3)(3(153+sqrt(11121))) + root(3)(3(153-sqrt(11121))))`

The Complex roots are:

`x_2 = 2+1/3(omega root(3)(3(153+sqrt(11121))) + omega^2 root(3)(3(153-sqrt(11121))))`

`x_3 = 2+1/3(omega^2 root(3)(3(153+sqrt(11121))) + omega root(3)(3(153-sqrt(11121))))`

where `omega = -1/2+sqrt(3)/2i = cos((2pi)/3)+i sin((2pi)/3)` is the primitive Complex cube root of unity.