Factoring Quartics (in quadratic form)

Question

I have some quartic polynomial I wish to factor. Here is an example:

$x^4 + x^2 + 1 $

I know the answer to this question

$ (x^2 + x + 1)(x^2 -x +1) $

We get these 2 irreducible quadratics. I am trying to see if there is some systematic way, a process to arrive at this, without having to figure out a pattern for this.

Answer 1

If you have this nice sort of symmetry there's a trick you can apply; divide through by $x^2, x\ne 0$ to get $$x^2 + 1 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 -1$$

Using the substitution $u=x + 1/x$ allows you to solve for the roots of the polynomial.

Something else you could do: take the polynomials $ax^2 + bx + c$ and $dx^2+ex + f$, multiply them together and solve for $a,b,c,d,e,f$. Note that you'll end up with 6 unknowns and 5 equations, so you'll have to choose one of the values (say $a=1$).

Answer 2

You can also use the differences of squares identity just as below: $$x^4+x^2+1=x^4+2x^2+1-x^2$$ The first three summands is conveniently $(x^2+1)^2$ and the equation can be rewritten into: $$(x^2+1)^2-x^2$$ and by differences of squares identity you would get your two quadratics.