## You are here

Homeexample of using Eisenstein criterion

## Primary tabs

# example of using Eisenstein criterion

For showing the irreducibility of the polynomial

$P(x)\;:=\;x^{5}\!+\!5x\!+\!11$ |

one would need a prime number dividing its other coefficients except the first one, but there is no such prime. However, a suitable substitution $x:=y\!+\!a$ may change the situation. Since the binomial coefficients of $(y\!-\!1)^{5}$ except the first and the last one are divisible by 5 and $11\equiv 1\;\;(\mathop{{\rm mod}}5)$, we try

$x\;:=\;y-1.$ |

Then

$P(y\!-\!1)\;=\;y^{5}\!-\!5y^{4}\!+\!10y^{3}\!-\!10y^{2}\!+\!10y\!+\!5.$ |

Thus the prime 5 divides other coefficients except the first one and the square of 5 does not divide the constant term of this polynomial in $y$, whence the Eisenstein criterion says that $P(y\!-\!1)$ is irreducible (in the field $\mathbb{Q}$ of its coefficients). Apparently, also $P(x)$ must be irreducible.

It would be easy also to see that $P(x)$ does not have rational zeroes.

## Mathematics Subject Classification

13A05*no label found*11C08

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections