# 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.

• Out of curiosity, what was the genesis of this equation, where did you run across it? Dec 14, 2014 at 17:08

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

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

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

• That's just amazing!!
– user196761
Dec 13, 2014 at 16:10
• Wow! How do you think of such things!? Dec 13, 2014 at 16:15
• Well, he is a Math God. . . Dec 13, 2014 at 16:15
• @KierenMacMillan Can you tell me what solution does maxima give? Jun 27, 2015 at 15:09
• Apologies. After doing some arithmetic, the maxima solution is the same as yours. My fault. I deleted my comment, and upvoted your answer. Jun 27, 2015 at 16:09

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

• I believe it will be more helpful, if you show, how you get this form. (And no, I didn't down-vote.) Dec 13, 2014 at 16:12

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

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

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

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

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

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

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

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

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

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

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

The $$7$$ roots of $$x$$ are:

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