= Solution
A <cubic polynomial> over a <field> is reducible if and only if it has a root in that field: a nontrivial factorization of degree three has a <linear polynomial> factor, and conversely a root supplies one by the <factor theorem>.
The <rational root theorem>, or part (b), shows that a rational root of this monic polynomial must be an integer. Such an integer must divide the constant coefficient $4$, so the only candidates are $\pm1,\pm2,\pm4$. Direct evaluation gives
$$
\begin{array}{c|rrrrrr}
a&1&-1&2&-2&4&-4\\\hline
h(a)&9&-1&20&-12&84&-76
\end{array}
$$
None vanishes. Hence there is no rational root, proving
$$
\boxed{t^3+4t+4\text{ is irreducible over }\mathbb Q.}
$$
The same no-root argument alone would not decide a <quartic polynomial>, which could factor as two <quadratic polynomials>; its sufficiency here uses degree three.
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