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

• amcavoy
In summary, the polynomial x^5+x+1 cannot be factored as it is already in its simplest form. However, it can be partially factored as (x^3-x^2+1)(x^2+x+1) in the reals. Using Cardano's formula, it can be factored into monomials over the complex numbers, but this is a difficult and tedious process. Overall, it is not possible or practical to fully factor this polynomial.
amcavoy
How can I factor the following polynomial?

$$x^5+x+1$$

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
$$(x^3-x^2+1)(x^2+x+1)$$=$$(x^5+x+1)$$=$$x(x^4+1)+1$$

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

But u did it...Congratulations !

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

## 1. What is the process for factoring x^5+x+1?

The first step in factoring x^5+x+1 is to check if there are any common factors among the terms. In this case, there are no common factors. Next, we can try to factor by grouping. However, since there are only three terms, this method is not applicable. The final step is to use the quadratic formula to find the roots of the polynomial. Unfortunately, in this case, the polynomial does not have any real roots. Therefore, x^5+x+1 is not factorable.

## 2. Can I use the difference of squares method to factor x^5+x+1?

No, the difference of squares method can only be used when there are two terms in the polynomial. In this case, there are three terms, so this method is not applicable.

## 3. Is it possible to factor x^5+x+1 using synthetic division?

No, synthetic division can only be used to divide a polynomial by a linear factor. In this case, the polynomial x^5+x+1 does not have any linear factors, so synthetic division cannot be used to factor it.

## 4. How do I know if a polynomial is prime?

A polynomial is considered prime if it cannot be factored into two polynomials with integer coefficients. In other words, if the polynomial does not have any factors other than 1 and itself, it is considered prime. In the case of x^5+x+1, since it cannot be factored, it is considered a prime polynomial.

## 5. Is there a general formula for factoring polynomials?

No, there is no general formula for factoring polynomials. However, there are various methods and techniques that can be used depending on the type of polynomial. It is important to first check for any common factors and then try different methods, such as grouping, difference of squares, or using the quadratic formula to find the roots, to factor the polynomial.

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