Irreducible factor decomposition This is a past exam question.
Decompose each of the following elements as a product of irreducible:
(a) $X^4+2 \in \mathbb{Z}_5[X]$
(b) $X^5+X \in \mathbb{Z}_2[X]$
(c) $X^5+4X^4-3X^3+X^2+7X+11 \in \mathbb{Q}[X]$

(a) I run over the value $0$ to $4$, and show this polynomial has no linear factors, but how do I show this polynomial also has no quadratic factor? Do I write down 
$$X^4+2 = (aX^2+bX+c)(a^\prime X^2+b^\prime X+c^\prime)$$
and show there are no such $a,b,c,a^\prime,b^\prime,c^\prime \in \mathbb{Z}_5$ Which is something that I don't want to do in the exam, will take quite a long time.
(b) $X=0,1$ is a root for the polynomial. and after some long divisions. $X^5+X=X(X+1)^4$
(c) $X^5+4X^4-3X^3+X^2+7X+11$ has no linear factors in $\mathbb{Z}_2[X]$. And I know all the irreducible polynomials in $\mathbb{Z}_2[X]$ with degree $\le$ 3. So I can use long division to see can they divide $X^5+4X^4-3X^3+X^2+7X+11$.

And for a more general question, how do I test the irreducibility for a polynomial with degree higher than 3. Since they might have quadratic factor or factor of higher power.
 A: Hope this answer can help people who are doing algebra.
(a) $f(X)=X^4+2 \in \mathbb{Z}_5[X]$
$f(0)=2,f(1)=3,f(2)=3,f(3)=3,f(4)=2 \Rightarrow f(X)$ has no linear factors in $\mathbb{Z}_3[X]$.
Hence if $f(X)$ is reducible, then it will have 2 quadratic factors for which they are monic.
$$X^4+2 = (X^2+bX+c)(X^2+b^\prime X+c^\prime)$$
Gives,
$$b = -b^\prime \space\space\space\space\space\space\space\space -(1)\\
c+c^\prime+bb^\prime=0 \space\space\space\space\space\space\space\space -(2)\\
b^\prime c+bc^\prime = 0 \space\space\space\space\space\space\space\space -(3)\\
cc^\prime=2 \space\space\space\space\space\space\space\space -(4)$$
Sub. $(1)$ into $(3)$, we get $-bc+bc^\prime=0 \Rightarrow b=0$ or $c=c^\prime$
If $b=0$, $(2) \Rightarrow c=-c^\prime$, $(4) \Rightarrow c^2 =2$ (contradiction)
If $c=c^\prime$, $(4) \Rightarrow c^2 =2$ (contradiction)
Therefore, $X^4+2$ is irreducible in $\mathbb{Z}_5[X]$
(b) $X=0,1$ is a root for the polynomial. and after some long divisions. $X^5+X=X(X+1)^4$ in $\mathbb{Z}_2[X]$
(c) $f(X)=X^5 + 4 X^4 - 3 X^3 + X^2 + 7 X + 11 \in \mathbb{Q}[X]$
$$\bar{f}(X) = X^5+X^3+X^2+X+1 \in \mathbb{Z}_2[X]$$
$\bar{f}(0)=\bar{f}(1)=1 \Rightarrow \bar{f}(X)$has no linear factors in $\mathbb{Z}_2[X]$.
So, $\bar{f}(X)$ must be decompose into $$\text{(polynomial of degree 2) * (polynomial of degree 3)}$$
There are only one irreducible polynomial of degree 2 in $\mathbb{Z}_2[X]$ which is $X^2+X+1$. By long division, we can show that $X^2+X+1 \nmid \bar{f}(X)$. Hence $\bar{f}(X)$ is irreducible in $\mathbb{Z}_2[X]$ also $deg(f(X))=deg(\bar{f}(X))$ (Note: this is same as saying the prime number 2 does not divide the highest order coefficient of $f(X)$), then $f(X)$ is irreducible in $\mathbb{Q}[X]$.
