Solve $x^7-5x^4-x^3+4x+1=0$ for $x$

Solve for $x$
$$x^7-5x^4-x^3+4x+1=0$$

This equation has been bugging me since the past few days. I have found, using the Rational Root Theorem that $x=1$ is a root of this equation. However, after dividing, I cannot solve the six degree equation thus generated. I have also tried factorizing the equation, but it's not working.

Solution 1:

Consider the identity

$(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$

Differentiating both sides w.r.t. $x$, we get,

$x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$

Now, the equation becomes,

$(x^4-4x-1)(x^3-1)=0$

$\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity)

or

$x^4-4x-1=0$

$\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$

$\implies (x^{2}+1)^{2}=2(x+1)^{2}$

$\implies \{x^{2}+1+\sqrt{2} (x+1)\}\cdot \{x^{2}+1-\sqrt{2} (x+1)\}=0$

$\implies x=\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2}$

or

$x=\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$

$\therefore x=1,\omega,\omega^2,\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2},\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$

Solution 2:

you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem