Question

How do you factor `x^12+8`?

Answer 1

Use some identities to find the quadratic factors of `x^12+8`

Explanation:

First notice that `x^12 >= 0` for any Real value of `x`. So `x^12+8 >= 8 > 0` for any Real `x` and this polynomial has no linear factors with Real coefficients.

We can find all of the quadratic factors as follows:

The sum of cubes identity can be written:

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

Hence:

`x^12+8=(x^4)^3+2^3`

`=(x^4+2)((x^4)^2-(x^4)(2)+2^2)`

`=(x^4+2)(x^8-2x^4+4)`

Notice that:

`(a^2-kab+b^2)(a^2+kab+b^2)=(a^4+(2-k^2)a^2b^2+b^4)`

In particular, if `k=sqrt(2)` then we find:

`(a^2-sqrt(2)ab+b^2)(a^2+sqrt(2)ab+b^2)=a^4+b^4`

Hence (putting `a=x` and `b=root(4)(2)`) we find:

`(x^4+2) = (x^2-root(4)(8)x+sqrt(2))(x^2+root(4)(8)x+sqrt(2))`

If instead we put `k=sqrt(3)` then we find:

`(a^2-sqrt(3)ab+b^2)(a^2+sqrt(3)ab+b^2) = a^4-a^2b^2+b^4`

So if we let `a=x^2` and `b=sqrt(2)` then we find:

`(x^4-sqrt(6)x^2+2)(x^4+sqrt(6)x^2+2) = x^8-2x^4+4`

Putting `k=sqrt(2+sqrt(3))`, `a=x` and `b=root(4)(2)` we find:

`(x^2-sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))`

`=x^4-sqrt(6)x^2+2`

Putting `k=sqrt(2-sqrt(3))`, `a=x` and `b=root(4)(2)` we find:

`(x^2-sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))`

`=x^4+sqrt(6)x^2+2`

Putting this all together, we find:

`x^12+8 = (x^2-root(4)(8)x+sqrt(2))(x^2+root(4)(8)x+sqrt(2))(x^2-sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2+sqrt(3))root(4)(2)x+sqrt(2))(x^2-sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))(x^2+sqrt(2-sqrt(3))root(4)(2)x+sqrt(2))`