How does one prove that $(x^4 - 5)$ and $(x^4 + 5)$ are irreducible over $\mathbb{Q}$? I have factorised $x^8 - 25$ into $(x^4 - 5)(x^4 + 5)$ but I'm unsure about how to go about showing that it is fully factorised. 
Thanks very much in advance.

A very hands on way since your polynomials are of degree $4$ (not very high degree). Suppose that $x^4-5$ is reducible over $\mathbb{Q}$ (the other polynomial the same argument). Then there exist nonconstant polynomials $p(x),q(x)$ with degree smaller than $4$ and with rational coefficients such that $x^4-5=p(x)q(x)$. There are two possibilities, either $p(x)$ is of degree $1$ and $q(x)$ is cubic or both are quadratic. Say $p(x)=x+a, q(x)=x^3+bx^2+cx+d$ with $a,b,c,d\in\mathbb{Q}$. Then upon multiplication we get
$$p(x)q(x)=(x+a)(x^3+bx^2+cx+d)=x^4+(a+b)x^3+(c+ab)x^2+(d+ac)x+ad$$
On the other hand
$$p(x)q(x)=x^4-5$$
This implies $a+b=0,c+ab=0,d+ac=0,ad=-5$ which yield $a=-b$ and consequently $c=b^2, d=b^3$ giving $ad=-bb^3=-b^4=-5$ or $b=\sqrt[4]{5}\notin\mathbb{Q}$. Hence a contradiction to our assumption $b\in\mathbb{Q}$. In the other case 
$$p(x)q(x)=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d)x^2+(ad+bc)x+bd$$
Again by looking at the coefficients we obtain $a+c=0, b+d=0,ad+bc=0$ and $bd=-5$. These imply $bd=-b^2=-5$ or $b=\sqrt{5}\notin\mathbb{Q}$. Again a contradiciton. For cases of higher degrees of course this method is laborious and you should refer to other ways like Eisenstein's theorems for polynomial reductions. 
