Reduction modulo p I am going to begin the Tripos part III at Cambridge in October (after going to a different university for undergrad) and have been preparing by reading some part II lecture notes. 
Here is an extract from some Galois theory notes:
'$f(X) = X^4 + 5X^2 − 2X − 3 = X^4 + X^2 + 1 = (X^2 + X + 1)^2$ (mod 2)
$f(X) = X^4 + 5X^2 − 2X − 3 =X^4 + 2X^2 + X = X(X^3 + 2X + 1)$ (mod 3) 
So f is irreducible, since f = gh implies deg g = 1 or deg g = 2, which is impossible by
reduction modulo 2 and 3, respectively.'
Could someone please explain the last sentence? Why is deg g=1 or 2 impossible by reduction modulo 2/3?
Thank you in advance.

Hint $\ 4$ is the only partition of $4$ having common refinements $2+2$ and $1+3$
