How to Factor This?

1. Apr 9, 2005

amcavoy

How can I factor the following polynomial?

$$x^5+x+1$$

Thanks for your help.

2. Apr 9, 2005

dextercioby

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

Daniel.

3. Apr 9, 2005

huan.conchito

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

Last edited: Apr 9, 2005
4. Apr 9, 2005

dextercioby

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

But u did it...Congratulations !

Daniel.

5. Apr 9, 2005

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.

6. Apr 10, 2005

Zurtex

Erm this is correct.

7. Apr 11, 2005

Data

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.

8. Apr 12, 2005

CRGreathouse

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.

9. Apr 13, 2005

dextercioby

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?