Question

How do you factor the expression `6x^4+24x^2-72x`?

Answer 1

`6x^4+24x^2-72x`

`=6x(x^3+4x-12)`

`=6x(x-x_1)(x-x_2)(x-x_3)`

where `x_1`, `x_2` and `x_3` are the zeros of the cubic as follows...

Explanation:

Firstly, all of the terms are divisible by `6x`, so we can separate that out as a factor:

`6x^4+24x^2-72x=6x(x^3+4x-12)`

We can factor the remaining cubic factor using Cardano's method:

Let `x = u + v` and solve `x^3+4x-12 = 0`

`u^3+v^3+(3uv+4)(u+v)-12 = 0`

Add the constraint `v = -4/(3u)` to eliminate the term in `(u+v)` and get:

`u^3-(4/(3u))^3-12 = 0`

Multiply through by `27u^3` to get:

`27(u^3)^2-324(u^3)-64 = 0`

Use the quadratic formula to find:

`u^3=(324+-sqrt(324^2+4*27*64))/54`

`=(324+-sqrt(111888))/54`

`=(324+-12sqrt(777))/54`

`=(162+-6sqrt(777))/27`

Since the square root is Real, this gives us a Real value for `u^3`.so since the derivation was symmetric in `u` and `v` we can use one of these roots for `u^3` and the other for `v^3` to find the Real root:

`x_1 = 1/3(root(3)(162+6sqrt(777)) + root(3)(162-6sqrt(777)))`

and Complex roots:

`x_2 = 1/3(omega root(3)(162+6sqrt(777)) + omega^2 root(3)(162-6sqrt(777)))`

`x_3 = 1/3(omega^2 root(3)(162+6sqrt(777)) + omega root(3)(162-6sqrt(777)))`

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

Then:

`x^3+4x-12 = (x-x_1)(x-x_2)(x-x_3)`