How to Factor This?

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  • #1
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How can I factor the following polynomial?

[tex]x^5+x+1[/tex]

Thanks for your help.
 

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  • #2
dextercioby
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U can't...It has only one real root and that's it...And that root is really ugly.

Daniel.
 
  • #3
took me a while to get it right
[tex](x^3-x^2+1)(x^2+x+1)[/tex]=[tex](x^5+x+1)[/tex]=[tex]x(x^4+1)+1[/tex]
 
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  • #4
dextercioby
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It doesn't give you too much,though...2 complex solutions out of 5.

But u did it...Congratulations ! :smile:

Daniel.
 
  • #5
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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
Zurtex
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eNathan said:
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.
huan.conchito said:
[tex](x^3-x^2+1)(x^2+x+1)=(x^5+x+1)[/tex]
Erm this is correct.
 
  • #7
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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.
 
  • #8
CRGreathouse
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huan.conchito said:
took me a while to get it right
[tex](x^3-x^2+1)(x^2+x+1)[/tex]=[tex](x^5+x+1)[/tex]=[tex]x(x^4+1)+1[/tex]
This is as far as the polynomial can be factored in [tex]\mathbb{Z}[x][/tex], 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 [tex]f=x^3-x^2+1[/tex] and substituting [tex]z=x-\frac13[/tex] we get [tex]f=z^3+z^2-\frac13z+\frac{1}{27}-z^2-\frac23z-\frac19+1=z^3-z+\frac{25}{27}[/tex].



Using Cardano's formula, we have [tex]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}}}[/tex]

which "simplifies" to

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

Well, I suppose this could be used to factor this into monomials over [tex]\mathbb{C}[x][/tex], but I'd really hate to actually do it. I like [tex](x^3-x^2+1)(x^2+x+1)[/tex] much better.
 
  • #9
dextercioby
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[tex] x^3-x^2+1=0[/tex], 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\} $
 

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