# 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 $11\equiv 1\pmod{5}$,  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 (http://planetmath.org/RationalRootTheorem).

Title example of using Eisenstein criterion ExampleOfUsingEisensteinCriterion 2013-03-22 19:10:14 2013-03-22 19:10:14 pahio (2872) pahio (2872) 5 pahio (2872) Example msc 13A05 msc 11C08