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The polynomial is Eisenstein at $2$: every nonleading coefficient is even, and its constant term $126$ is not divisible by $4$. Therefore it is irreducible over $\mathbb{Q}$ andBy part (a), the number of primes above a rational prime is the number of irreducible factors over the corresponding $p$-adic field.

We use the Newton polygon theorem in this form: a segment of slope $-r$ and horizontal length $d$ accounts for $d$ roots of valuation $r$, counted with multiplicity, in an algebraic closure with $v_p(p)=1$. We also use the denominator criterion for Newton polygon irreducibility: if all roots have valuation $a/n$ with $\gcd (a,n)=1$, any monic factor of degree $d$ has constant-term valuation $da/n\in \mathbb{Z}$, hence $n\mid d$.

At $3$, the nonzero coefficient points areThe lower hull is the single segment from $(0,2)$ to $(5,0)$, with slope $-2/5$. All five roots therefore have valuation $2/5$. A proper factor of degree $d$ would require $5\mid d$, which is impossible for $1\le d<5$. Thus $g$ is irreducible over ${\mathbb{Q}}_3$, givingThe completion has ramification index $5$ and residue degree $1$: the value $2/5$ forces its ramification index to be divisible by $5$, and its degree is $5$.

At $7$, the nonzero coefficient points areThe lower hull consists of a length-three segment of slope $-1/3$ and a length-two horizontal segment. Thus three roots have valuation $1/3$, and two have valuation zero. Reduction givesBoth $1$ and $-1$ are simple roots, so the Hensel lemma gives two distinct linear factors over ${\mathbb{Q}}_7$. The remaining cubic accounts for all the valuation-$1/3$ roots and is irreducible by the same denominator argument. Hence the local factor degrees are $1,1,3$, andThe two linear factors give $(e,f)=(1,1)$, and the cubic gives $(e,f)=(3,1)$. These data sum to degree five, as required. This computes the local prime decomposition of T5 plus 6T3 plus 252T2 plus 126. Counting distinct factors merely modulo $3$ would not justify the result, since that reduction is $T^5$.