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

Question

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.

Answer 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

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

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

Answer 2

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

Answer 3

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

$\implies(x^3 + 1)(x^4 + 1) = 0$

$\implies(x + 1)(x^2 - x + 1)(x^4 + 2x^2 + 1 - 2x^2) = 0$

$\implies(x + 1)(x^2 - x + 1)((x^2 + 1)^2 - 2x^2) = 0$

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

$ x^2 - x + 1 = 0$

$\implies x = \frac{1 \pm \sqrt{1 - 4\times1}}2 = \frac{1 \pm i \sqrt 3}2$

$ x^2 + \sqrt 2x + 1 = 0$

$\implies x = \frac{-\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{-\sqrt 2 \pm i \sqrt 2}2$

$ x^2 - \sqrt 2x + 1 = 0$

$\implies 4x = \frac{\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{\sqrt 2 \pm i sqrt 2}2$

The $7$ roots of $x$ are:

$x = -1, \frac{1 \pm i \sqrt 3}2, \frac{-\sqrt 2 \pm i \sqrt 2}2, \frac{\sqrt 2 \pm i \sqrt 2}2$