Elegant way to find the roots of a Quartic I need to find the roots of this quartic: $y=x^4+6x^3+11x^2+6x-24024$ in order to be able to solve my problem.
I'm wondering if there's a nice easy way to find its roots (such as substituting $x$ with something). I do know the Rational Root Theorem, but I feel like listing out the factors of $24024$ would take too long.

You may notice that
$$ x^4+6x^3+11x^2+6x-24024 = (x^2+3x+1)^2-24025 = (x^2+3x+1)^2-155^2 $$
by looking for the square of a second degree polynomial that matches the coefficients pattern $1,6,11$. It goes so smooth that the identity $a^2-b^2=(a-b)(a+b)$ also provides a factorization:
$$\boxed{ x^4+6x^3+11x^2+6x-24024 = \color{blue}{(x-11)(x+14)}\color{green}{(x^2+3x+156)}.}$$
The fact that $24024$ is exceptionally close to an integer square and $x^4+6x^3+11x^2+6x+1$ is a palyndromic polynomial (hence $x^2$ times a quadratic polynomial in the variable $x+\frac{1}{x}$) and a square leads to an alternative, equivalent solution.
