$6x^5+14x^3-21x+35$ is irreducible in $\mathbb{Q}[x]$ I am trying to show that $p(x)=6x^5+14x^3-21x+35$ is irreducible in $\mathbb{Q}[x]$. 
I would like to be able to use Eisenstein's Criterion which states:
Let $P$ be a prime ideal of the integral domain $R$ and let $f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ be a polynomial in $R[x]$, where $n \geq 1$. Suppose that $a_{n-1}, \ldots, a_1 \in P$ and suppose that $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible in $R[x]$. 
The problem is that the polynomial $p(x)$ that I am given does not have a leading coefficient of 1. Is there another approach I can take? I would like to be able to write a self-contained proof, if possible.
 A: You can get rid of the coefficient $6$ by multiplying your equation by the quantity $6^4$, then performing the substitution $y = 6x$; this will give you a MONIC polynomial in y with integer coefficients.  Eisenstein now applies.
A: Because $p(x)\in\Bbb{Z}[x]$ is primitive, by Gauss' lemma it suffices to show $p(x)$ is irreducible in $\Bbb{Z}[x]$. Suppose toward a contradiction that $p(x)$ is reducible, say $p(x)=f(x)g(x)$ with $\deg f,\deg g>0$. Reducing modulo $7$ yields two polynomials $\overline{f}(x),\overline{g}(x)\in\Bbb{F}_7[x]$ such that $\overline{f}(x)\overline{g}(x)=\overline{p}(x)=\overline{6}x^5$. This means $\overline{f}(x)=ax^m$ and $\overline{g}(x)=bx^n$ for some $a,b\in\Bbb{F}_7$ and $m,n\in\Bbb{N}$. Because the product of the leading coefficients of $f$ and $g$ is $6$, the leading coefficients of $f$ and $g$ are not divisible by $7$, so 
$$m=\deg\overline{f}=\deg f>0\qquad\text{ and }\qquad n=\deg\overline{g}=\deg g>0.$$
This means the constant coefficients of $f(x)$ and $g(x)$ are divisible by $7$, and so the constant coefficient of $p(x)=f(x)g(x)$ is divisible by $7^2$. But $35$ is not divisible by $7^2$, a contradiction. Hence $p(x)$ is irreducible in $\Bbb{Z}[x]$, and by Gauss' lemma it is irreducible in $\Bbb{Q}[x]$.
A: Take $R=\Bbb{Z}[\frac16]$, $P=7R$, and $a_4=0$, $a_3=2\cdot\frac{1}{6}\cdot7$, $a_2=0$, $a_1=-3\cdot\frac{1}{6}\cdot7$ and $a_0=5\cdot\frac{1}{6}\cdot7$.
Verify that $5\cdot\frac16\cdot7\notin7^2R$ to find that 
$$f(x)=x^5+\frac{14}{6}x^3-\frac{21}{6}x+\frac{35}{6},$$
is irreducible in $R[x]$, hence so is your polynomial $6f(x)$ because $6$ is a unit in $R$. Then certainly your polynomial is also irreducible in $\Bbb{Z}[X]$ because $\Bbb{Z}\subset R$, and so by Gauss' lemma it is irreducible in $\Bbb{Q}[X]$.

Needless to say this argument holds for any polynomial whose leading coefficient is not in $P$, and it is worth verifying that Eisenstein's criterion works not only for monic polynomials, but for all polynomials satisfying $a_n\notin P$, where $a_n$ is the leading coefficient.
A: Use p=7. 7 does not divide 6 it divides 14,21,and 35 but 7^2=49 does not divide 35. Hence, the polynomial is Eisenstein at p=7 and thus is irreducible in $\mathbb{Q}[x]. 
Now, this is an equivalent formulation of the Eisenstein Criterion you stated above. 
