Solve the polynomial $x^6+x^5+x^4-2x^3-x^2+1=0$ in exact form

Question

Try to reduce the degree of the polynomial $$P(x)=x^6+x^5+x^4-2x^3-x^2+1$$ by algebraic ways and find the possible solution method to $P(x)=0$.

The source of the problem comes from a non-english algebra precalculus workbook. It is a workbook for those who want to be more interested in mathematics.

WolframAlpha couldn't solve the polynomial in exact form. Only approximate solutions are possible.

Rational root theorem shows that there are no rational roots. So, first method failed.

But, I tried different ways.

$$\frac {P(x)}{x^2}=x^3+x^2+x-2-\frac 1x+\frac 1{x^3}=\bigg(x^3+\frac 1{x^3}\bigg)+\bigg(x+\frac 1x\bigg)+x^2-2=0$$

or

$$\frac {P(x)}{x^2}=x^4+x^3+x^2-2x-1+\frac {1}{x^2}=\bigg(x^2+\frac 1{x^2}\bigg)+(x^4+x^3-2x-1)$$

I think that the trick $x+\frac 1x=t$ doesn't work here.

At the end I tried factoring the Polynomial but here are $2$ possibilities to do it

$$P(x)=(x^2+ax+b)(x^4+cx^2+dx+e)$$

or

$$P(x)=(x^3+ax^2+bx+c) (x^3+dx^2+ex+f)$$

But which of these methods will work, it is not clear.

Since the polynomial has no real roots, the second factorization certainly doesn't work. And since the real part of the roots are still a mess, I doubt the existence of a clear factorization. — Zerox, Sep 18 '22 at 16:47
See this question. The Galois group here is $D_6$, which is solvable. Again, this is not "algebra-precalculus". — Dietrich Burde, Sep 18 '22 at 16:48

score 4 · Accepted Answer · answered Sep 18 '22 at 17:24

The trick $x+\frac1x=t$ requires the polynomial to be a palindrome, so that is clearly not the case here. However that's not the only substitution possible to simply. Here, lets try something similar to what you have already done, persisting a bit more:

$ \begin{align} \frac{P(x)}{x^4} &= x^2+x+1-\frac2x-\frac1{x^2}+\frac1{x^4} \\ &= \left(x^2-\frac2x+\frac1{x^4}\right)+\left(x-\frac1{x^2} \right)+1\\ &= \left(x-\frac1{x^2} \right)^2+\left(x-\frac1{x^2} \right)+1\\ &=u^2+u+1 \tag{1} \end{align} $

where $u = x-\dfrac1{x^2} \tag{2}$

Now clearly $u$ is solvable as a quadratic from $(1)$, and for both values of $u$, the equation $(2)$ gives a cubic to solve for $x$. As both quadratics and cubics admit solutions in exact algebraic form (though using cumbersome radicals perhaps), we are done.

In other words, you have shown that the sextic splits in $\mathbb Q(\zeta_6)$ ($\zeta^2-\zeta+1=0$):$$x^3+\zeta x^2-1=0$$or its conjugate. — Parcly Taxel, Sep 18 '22 at 18:11

Solve the polynomial $x^6+x^5+x^4-2x^3-x^2+1=0$ in exact form

1 Answers1