Using the natural homomorphism $\mathbb Z$ to $\mathbb Z_5$ 
Prove that $x^4+10x^3+7$ is irreducible in $\mathbb Q[x]$ by using the natural homomorphism from $\mathbb Z$ to $\mathbb Z_5$.

So I would assume we should rewrite our polynomial, maybe as $(x + 10) x^3 + 7$? Then in terms of $\mathbb Z_5, (x+2\cdot 5)x^3+(5+2)\equiv (x+0)x^3+2=x^4+2$. Although I am not sure of this because I know that $F_5$ has no zero divisors, so maybe 10 cannot be canceled. Also, this didn't use the natural homomorphism, so this cannot be right. Could I have some help?

The other answers do not address why there cannot be a quadratic factor.  I don't know if there's a clever thing I'm missing, but here's one way to see.  Assume the quadratic splits over $\mathbb{Q}$ as 
$$
x^4 + 2 = (x^2+ax+b)(x^2+cx+d) 
$$
Then, passing to $\mathbb{F}_5$ we get the equations
$$
\begin{eqnarray}
d+ac+b = 0 \\
a+c = 0 \\
ad+bc=0 \\
bd = 2
\end{eqnarray}
$$
From which we get
$$
\begin{eqnarray}
a(d-b)=0
\end{eqnarray}
$$
So either $a=0$ or $b=d$.  If $b=d$ then $b^2 = 2$, but 
$$
\begin{eqnarray}
0^2=0\\
1^2=1\\
2^2=4\\
3^2=4\\
4^2=1
\end{eqnarray}
$$
so, $b^2=-2$ is not possible.  Therefore $a=0$.  But then from $d+ac+b=0$ we get $d=-b$, so $-b^2=2$, and $-2$ is not a square either.  Therefore we can't factor $x^4+2$ in to quadratic polynomials over $\mathbb{F}_5$.  Since a factorization over $\mathbb{Q}$ would descend to $\mathbb{F}_5$, there is also no factorization in to quadratics over $\mathbb{Q}$. 
