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?
I will do all arithmetic in $\mathbb{Z}_5$. I will repeatedly use the fact that multiplication is commutative in this field. Consider $$ 0=x^4+2 $$ We see that $x \ne 1$, and $x \ne 2$ since $2^4 = 1$. Also, $x \ne 3$, since $3^4=1$. Finally $x \ne 4 = -1$, since $-1^4=1$. So there is no root to this polynomial in $\mathbb{Z}_5[x]$.
Hence we cannot write $$ x^4 + 2 = (a x + b)(cx^3+dx^2+ex+f) $$ because $x = a^{-1}(-b)$ would be a root.
Now let's check if $$ x^4+2 = (ax^2+ex+b)(cx^2+fx+d) $$ makes sense.
First, the quartic term implies $$ ac=1. $$
Second, the constant term implies that $$ bd=2. $$
Third, the cubic term implies that $$ af+ce = 0 $$ Fourth, the coefficient on $x$ implies that $$ bf+de=0\\ $$ Writing these last two equations as a system, we have $$ \left[ \begin{array}{cc} c & a \\ d & b \\ \end{array} \right] \left[ \begin{array}{c} e \\ f \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ \end{array} \right] $$ I claim that matrix is not singular. If it were, then its determinant would be zero. Then we could write: $$ cb-ad=0\\ cb=ad\\ acb=a^2d\\ b=a^2d \\ bd=a^2d^2 \\ 2=(ad)^2 \\ $$ and $2$ is not the square of any number in $\mathbb{Z}_5$. Hence $$ e = f = 0 $$ Considering the coefficient on $x^2$, we see that $$ ad+bc+ef=0\\ ad+bc=0\\ c^{-1}d+bc=0\\ d+bc^2=0\\ d=-bc^2 $$ Since $bd=2$, we may write $$ 2=bd=-b^2c^2=-(bc)^2\\ (bc)^2=-2=3 $$ There is no element of $\mathbb{Z}_5$ whose square is $3$.
So the polynomial is not reducible over $\mathbb{Z}_5[x]$.
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 $\mathbb{Q}$.
What you're missing about the "natural homomorphism" is:
So if it doesn't factor over $\mathbb{F}_5$, it doesn't factor over $\mathbb{Q}$.
There are a number of ways to show that $x^4 + 2$ is irreducible in $\mathbb{F}_5$. If $E$ is the splitting field, then $E / \mathbb{F}_5$ is a Kummer extension, since $\mathbb{F}_5$ has a fourth root of unity and you're adjoining a fourth root of an element of $\mathbb{F}_5$.
Even if you don't know about Kummer extensions, you can infer that $E$ has a sixteenth root of unity.
In any case, you can then deduce that $E = \mathbb{F}_{5^4}$. And there are a variety of ways to use that fact to show $x^4 + 2$ irreducible.
Exercise: To be sure you understand the overall argument, the same approach (with a little bit more work at the end) can be used to determine the degrees of the factors of $x^4 + 1$, and even what the factors are.
