I stumbled upon this kind of problem and I really can't get the hang of it. Will anyone please outline the way to solve it?
Determine for which of the first $p > 0$ values the polynomial $f = 42x^4+21x^3-x+1 \in \mathbb Z_p$ is monic and has degree 3. Then factor it as product of irreducibile polynomials in the polynomial rings found.
On
A polynomial has degree $n$ when the largest power of $x$ occurring in it that has a non-zero coefficient is $x^n$. For example, the polynomial $x^2+5x+1$ has degree 2, but so does the polynomial $0x^3+3x^2-x+4$.
The coefficient of this largest power of $x$ is called the leading coefficient of the polynomial. So, the leading coefficient of $x^2+5x+1$ is $1$, while the leading coefficient of the polynomial $0x^3+3x^2-x+4$ is $3$.
A polynomial is monic when its leading coefficient is $1$.
When does the polynomial $42x^4+21x^3-x+1$ have degree $3$? Only when $42=0$, and $21\neq 0$. When is it both of degree $3$ and also have leading coefficient $1$?
Hint $\ $ If $\rm\:f\ mod\ p\:$ is cubic then $\rm\:p\:|\:42,\ p\nmid 21\:$ hence $\rm\:p = \ldots$ Further, mod this $\rm\:p\:$ we see $\rm\:f\:$ has no roots, so $\rm\:f\:$ is an irreducible cubic. The point of this is that it implies that if $\rm\:f\:$ factors over $\Bbb Q\,$ then it must split as a linear times a cubic. Thus to show $\rm\:f\:$ is irreducible over $\Bbb Q\,$ it suffices to show it has no root, e.g. by using the Rational Root Test.
Edit $\ $ The OP later clarified that factorization over $\Bbb Q$ is not needed, so the second half of my answer is not needed. But I'll leave it since it may still prove of interest. In fact this is the way some polynomial factorization algorithms work: by deducing constraints on the degrees of possible factors from factorizations mod $\rm p$ for various primes $\rm p.$