How to Factor x^5+x+1 - Get Help Now!

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The polynomial x^5 + x + 1 cannot be factored into simpler polynomials over the reals, as it has only one real root and several complex roots. Some users noted that while it can be expressed as a product of (x^3 - x^2 + 1)(x^2 + x + 1), this does not simplify the polynomial significantly. Discussions also touched on the difficulty of factoring polynomials in general, with references to Cardano's method for cubic equations. Ultimately, the consensus is that x^5 + x + 1 is in its simplest form for real number factoring. The conversation highlights the challenges of polynomial factorization in advanced mathematics.
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How can I factor the following polynomial?

x^5+x+1

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

Daniel.
 
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
 
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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.
 
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.
 
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:
(x^3-x^2+1)(x^2+x+1)=(x^5+x+1)

Erm this is correct.
 
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.
 
huan.conchito said:
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.
 
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\} $
 

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