How to factor x^5+x+1

[amcavoy]

How can I factor the following polynomial?

$$x^5+x+1$$

Thanks for your help.

 

[dextercioby]

U can't...It has only one real root and that's it...And that root is really ugly.

Daniel.

 

[huan.conchito]

took me a while to get it right
$$(x^3-x^2+1)(x^2+x+1)$$=$$(x^5+x+1)$$=$$x(x^4+1)+1$$

 

[dextercioby]

It doesn't give you too much,though...2 complex solutions out of 5.

But u did it...Congratulations !

Daniel.

 

[eNathan]

like dextercioby said, concerning (x^5 + x + 1) it cannot be factored. Some things are already in their simplest form. X^2 is one monomial, x is one monomial, and 1 is one monomial. There are no like terms, thus they cannot be factored.

 

[Zurtex]

like dextercioby said, concerning (x^5 + x + 1) it cannot be factored. Some things are already in their simplest form. X^2 is one monomial, x is one monomial, and 1 is one monomial. There are no like terms, thus they cannot be factored.

$$(x^3-x^2+1)(x^2+x+1)=(x^5+x+1)$$

Erm this is correct.

 

[Data]

like dextercioby said, concerning (x^5 + x + 1) it cannot be factored. Some things are already in their simplest form. X^2 is one monomial, x is one monomial, and 1 is one monomial. There are no like terms, thus they cannot be factored.

Every real polynomial can be factored to a product of linear and quadratic functions over the reals. Of course, in almost every case this is a very difficult thing to do.

 

[CRGreathouse]

took me a while to get it right
$$(x^3-x^2+1)(x^2+x+1)$$=$$(x^5+x+1)$$=$$x(x^4+1)+1$$

This is as far as the polynomial can be factored in $$\mathbb{Z}[x]$$, unless I'm mistaken. Good job! I'm going to try to Cardano it, just to see if I can still do that...

c=1
d=-\frac{25}{27}

Working on $$f=x^3-x^2+1$$ and substituting $$z=x-\frac13$$ we get $$f=z^3+z^2-\frac13z+\frac{1}{27}-z^2-\frac23z-\frac19+1=z^3-z+\frac{25}{27}$$.



Using Cardano's formula, we have $$z=\sqrt[3]{-\frac{25}{54}+\sqrt{\frac{5^4}{2^23^6}-\frac{1}{3^3}}}+\sqrt[3]{-\frac{25}{54}-\sqrt{\frac{5^4}{2^23^6}-\frac{1}{3^3}}}$$

which "simplifies" to

$$z=\sqrt[3]{\frac{\sqrt{517}-25}{54}}-\sqrt[3]{\frac{\sqrt{517}+25}{54}}$$.

Well, I suppose this could be used to factor this into monomials over $$\mathbb{C}[x]$$, but I'd really hate to actually do it. I like $$(x^3-x^2+1)(x^2+x+1)$$ much better.

 

[dextercioby]

$$x^3-x^2+1=0$$, Solution is : $\left\{ x=-\sqroot{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }-\frac 1{9\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }}+\frac 13\right\} ,\allowbreak \left\{ x=\frac 12\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }+\frac 1{18\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }}+\frac 13+\frac 12i\sqrt{3}\left( -\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }+\frac 1{9\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }}\right) \right\} ,\allowbreak \left\{ x=\frac 12\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }+\frac 1{18\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }}+\frac 13-\frac 12i\sqrt{3}\left( -\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }+\frac 1{9\root{3}\of{\left( \frac{25}{54}+\frac 1{18}\sqrt{69}\right) }}\right) \right\} $