# example of using Eisenstein criterion

For showing the irreducibility (http://planetmath.org/IrreduciblePolynomial2) 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 $x:=y+a$ may change the situation. Since the binomial coefficients^{} of
${(y-1)}^{5}$ except the first and the last one are divisible by 5 and $11\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod5)$, we try

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

Then

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

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 (http://planetmath.org/RationalRootTheorem).

Title | example of using Eisenstein criterion |
---|---|

Canonical name | ExampleOfUsingEisensteinCriterion |

Date of creation | 2013-03-22 19:10:14 |

Last modified on | 2013-03-22 19:10:14 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 13A05 |

Classification | msc 11C08 |