Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to Factor This?

  1. Apr 9, 2005 #1
    How can I factor the following polynomial?

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

    Thanks for your help.
     
  2. jcsd
  3. Apr 9, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

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

    Daniel.
     
  4. Apr 9, 2005 #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]
     
    Last edited: Apr 9, 2005
  5. Apr 9, 2005 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

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

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

    Daniel.
     
  6. Apr 9, 2005 #5
    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.
     
  7. Apr 10, 2005 #6

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Erm this is correct.
     
  8. Apr 11, 2005 #7
    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.
     
  9. Apr 12, 2005 #8

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  10. Apr 13, 2005 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

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

Have something to add?