I am looking for the algebraic solution (not graphical) of the following equation.
$$2x^6+4x^5+x^4+4x^3+2x^2+1=0$$
According to the definition given in Wikipedia, this polynomial is not a palindromic polynomial, unfortunately.
But, a math student claims that the above polynomial has an algebraic solution. Our highschool teachers couldn't solve this.
Dividing each side by $x^4$, I get
$$2x^2+4x+1+\frac{4}{x}+\frac {2}{x^2}+\frac {1}{x^4}=0$$
So
$$2\bigg(x^2+\frac{1}{x^2}\bigg)+4\bigg(x+\frac {1}{x}\bigg)+\bigg(1+\frac{1}{x^4}\bigg)=0$$
I know that the property $$x^2+\frac {1}{x^2}=\bigg(x+\frac {1}{x}\bigg)^2-2$$
But the term $1+\frac {1}{x^4}$ violates the symmetry. What can I do from here?
Best Answer
With the help of Sagemath (another Computer Algebra System), I have found that the given polynomial $p$ factors over $\mathbb{Q}(\sqrt{2})$ as: $$p(x) = 2 \left(x^3 + \left(-\frac{\sqrt{2}}{2} + 1\right) x^2 - \frac{\sqrt{2}}{2} + 1 \right) \left( x^3 + \left(\frac{\sqrt{2}}{2} + 1\right) x^2 + \frac{\sqrt{2}}{2} + 1\right).$$ Now the roots of both of the cubic factors can be solved using the Cardano equation.
In retrospect, given the form of the factorization of $p$, you could put together a reverse-engineered type of solution for how somebody lucky might have come up with it: note that $$2 (x^3+x^2+1)^2 = 2x^6 + 4x^5 + 2x^4 + 4x^3 + 4x^2 + 2.$$ Therefore, $$p(x) = 2(x^3+x^2+1)^2 - (x^4 + 2x^2 + 1) = 2(x^3+x^2+1)^2 - (x^2+1)^2.$$