Another equivalent definition is $f\in k[X]$ is irreducible if $f=gh$ implies $g$ or $h$ is a unit in $k$ (degree zero).
Recall that in $\Z$ (or any integral domain) reducible element is a product of two non units, for example $20=4\cdot 5$ is reducible but $5=1\cdot 5$ is irreducible since $1$ is a unit. In particular primes are irreducible.

Examples

$2X\in\Q[X]$ is irreducible but it is reducible in $\Z[X]$ since $2$ is not a unit in $\Z$.
$f(X)= X^2-2\in\Q[X]$ is irreducible, but is reducible in $\mathbb{R}[X]$ as $(X-\sqrt{2})(X+\sqrt{2})$
$f(X)= X^2+1\in\Q[X]$ or $\mathbb{R}[X]$ is irreducible, but is reducible in $\C[X]$ as $(X-\ci)(X+\ci)$

Plug in $X=a/b$ and clear denominators to get $$c_na^n+c_na^{n-1}b+\ldots+c_1ab^{n-1}+c_0b^n=0,$$ since, $a$ and $b$ divide both sides of the equation, it gives $a|c_0$ and $b|c_n$.

According to the above proposition the polynomial $f(X)=X^3-5X+1\in\Q[X]$ can have only two possible roots $\pm 1$ but $f(1)=-3$ and $f(-1)=5$, thus $f(X)$ is irreducible.

Proof In $\Q[X], f=gh$, clear denominators to get $mnf=g_1h_1\in \Z[X]$ (here $m$ clears denominators of $g$ and $n$ clears denominators of $h$). We get rid of the prime factors of $mn$ by taking $\mod p$. $$0=\wbr{g_1}\wbr{h_1}\text{ in }\F_p[X]$$ $p$ must divide all coeff. of $g_1$ or $h_1$. Say it divides $g_1$, ( that is $g_1=pg_2$)$$\frac{mn}{p}f=g_2h_1\in\Z[X].$$ Continuing, remove all prime factors of $mn$ to get a factorization of $f$ in $\Z[X]$

Proof by contradiction. Suppose there exists a reducible $f(X)\in\Z[X]$ such that $\deg\wbr{f}(X)=\deg f(X)$ and $\wbr{f}(X)$ is irreducible in $\F_p[X]$, then necessarily $\wbr{f}(X)$ is reducible in $\F_p[X]$, because $\wbr{f}(X)=\wbr{g}(X)\wbr{h}(X)$, a contradiction.
In contrapositive terms, $f=gh$ implies $ \wbr{f}=\wbr{g}\wbr{h}$ if leading coeff of $f$ is not divisible by $p$. Thus reducibility of $f\in \Z[X]$ leads to reducibility in $\F_p[X]$.

Let $f=3X^3-5X^2+2X+7\in\Q[X]$ reducing $\mod 2$ gives $$\wbr{f}=X^3+X^2+1\in\F_2[X]$$ but $\wbr{f}(X)$ has no roots in $\F_2$ as $\wbr{f}(0)=1=\wbr{f}(1)$ and $\deg \wbr{f}(X)=\deg f(X)$, thus ${f}(X)\in\Z[X]$ is irreducible.

Proof by contradiction. Suppose, $f$ factors in $\Z[X]$ as $$(b_0X^r+\ldots+b_r)(c_0X^s+\ldots+c_s)$$ Then, $p|a_n$ implies $p|b_rc_s$, but $p^2\nmid b_rc_s$. Thus $p$ must divide only one of $b_r,c_s$. Say it divides $b_r$ $$a_{n-1}=b_rc_{s-1}+b_{r-1}c_s$$ Since $p|a_{n-1}, p|b_r, p\nmid c_s$ so $p|b_{r-1}$. Continuing this way $p|b_0,b_1,b_2,\ldots,b_r$ since $r<n$. But, this will contradict the condition that $p\nmid a_0.$

$$\Phi_p(X)=\frac{X^p-1}{X-1}=X^{p-1}+X^{p-2}+\ldots+X+1 \qquad\text{is irreducible}$$ If $\Phi_p(X)=g(X)h(X)$ then $ \Phi_p(X+1)=g(X+1)h(X+1)$, but $\Phi_p(X+1)$ is irreducible by Eisentein's Criterion as seen below (leading to a contradiction). $$\Phi_p(X+1)=\frac{(X+1)^p-1}{(X+1)-1}=X^{p-1}+pX^{p-2}+\ldots+p\qquad\text{is irreducible}$$ Hence, our assumption $\Phi_p(X)=g(X)h(X)$ is incorrect.