- #1

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How can I factor the following polynomial?

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

Thanks for your help.

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

Thanks for your help.

- Thread starter amcavoy
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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.

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

Thanks for your help.

- #2

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

Daniel.

Daniel.

- #3

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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]

[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:

- #4

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But u did it...Congratulations !

Daniel.

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- #6

Zurtex

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eNathan said:

Erm this is correct.huan.conchito said:[tex](x^3-x^2+1)(x^2+x+1)=(x^5+x+1)[/tex]

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- #8

CRGreathouse

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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...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]

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.

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