Does the Rational Root Theorem ever guarantee that a polynomial is irreducible? I know that the Eisenstein's criterion can guarantee it (when applicable), but what about this?
Also, is there another test that you can use to test irreducibility besides these two? For example, $x^4+2x^2+49$, over the rationals?
 A: With respect to your specific question (and quartics in general), the RRT doesn't solve the problem but it can help. Once you verify that your quartic has no roots, the only way it can factor is as a product of quadratic polynomials. Additionally, you can guarantee that the quadratic equations are monic and have integer coefficients. Observing that the constant terms multiply to $49$, we have four options:
$$x^4 + 2x^2 + 49 = (x^2 + ax + 7)(x^2 + bx + 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax - 7)(x^2 + bx - 7)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax + 1)(x^2 + bx + 49)$$
$$x^4 + 2x^2 + 49 = (x^2 + ax -1)(x^2 + bx - 49)$$
The vanishing of the cubic term gives $a = -b$ in each scenario, and the vanishing of the $x$ term eliminates scenarios 3 and 4. Comparing the $x^2$ terms in the first two scenarios gives:
$$14 - a^2 = 2$$
$$-14 - a^2 = 2$$
And it is easily seen that neither of these have solutions in the integers.
A: Using the rational root theorem you can tell if a given polynomial with integer coefficients has rational roots.
If the degree of the polynomial is greater than $3$ this theorem tells you nothing. For instance consider $(x^2-2)(x^2+2)=x^4-4$ which doesn't have rational roots, but is reducible over $\Bbb Q$.
If the degree of a given polynomial $f(x)$ is $1,2$ or $3$, then, if $f(x)$ is reducible, its degree is either $2$ or $3$ (since linear polynomials are irreducible). Therefore $f(x)=g(x)h(x)$ for some polynomials with rational coefficients and degrees greater than $0$. It follows that the degree of either $g(x)$ or $h(x)$ is $1$. Therefore $f(x)=p(x)(x-r)$, for some polynomial with rational coefficients $p(x)$ and for some $r\in \Bbb Q$. It follows that $f(r)=0$ and therefore $f(x)$ has a rational root.
Edit note: My previous reasoning didn't answer the question as noted by @Math Gems (thanks).
A: Other people (so far) are posting answers that warn against the use of the rational roots theorem for irreducibility above degree 3, but it is worth pointing out that there is a situation in higher degree where lack of a root actually does imply irreducibility.
Theorem:  Let $K$ be a field and $p$ be a prime number. For $a \in K^\times$, if the polynomial $X^p - a$ has no root in $K$ then $X^p - a$ is irreducible in $K[X]$.  Equivalently, if $a$ doesn't have a $p$th root in $K$ then $X^p - a$ is irreducible in $K[X]$. 
The theorem is true for all fields and prime numbers, even if $p$ is the characteristic of $K$.
A: Over a field, cubics and quadratics are irreducible iff they have no rational root, since if they are reducible they must have a linear factor, hence a rational root. 
But polynomials of degree $> 3$ can be reducible with all factors of degree $\ge 2$ and, thus, with no rational root. There are, however, some cases where irreducibility of higher degree polynomials can be reduced to root testing, e.g. this classic result:
Theorem $\ $ Suppose $\rm\:c\in F\:$ a field,  and $\rm\:0 < n\in\mathbb Z\:.$
$\rm\quad x^n - c\ $ is irreducible over $\rm\:F \iff c \not\in F^p\:$ for all primes $\rm\:p\: |\: n\:$ and $\rm\ c\not\in -4\:F^4\:$ when $\rm\: 4\ |\ n\:. $
A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.
Another method uses: $\rm\, p(x)\, $ is prime (irreducible) if it assumes a prime value
for large enough $\rm\, |x|.\: $ As an example, Polya-Szego popularized Arthur Cohn's irreduciblity test, that $\rm\, p(x) \in {\mathbb Z}[x]\,$ is prime if  $\rm\, p(b)\, $
notates a prime in radix $\rm\,b\,$  (so necessarily $\rm\,0 \le p_i < b),\:$
e.g. $\rm\,f(x) = x^4\! + 6 x^2\! + 1\,$ factors $\rm\,mod\ p\,$ for all primes $\rm\,p,\,$ yet  $\rm\,f(x)\,$ is prime since  $\rm\,f(8) = 10601\,$ octal $= 4481$ is prime. Note: Cohn's test fails if, in radix $\rm\,b,\,$ negative digits are allowed, e.g.
$\rm\ f(x) = x^3\! - 9 x^2\! + x-9 = (x-9)(x^2 + 1)\ $ but $\rm\,f(10) = 101\,$ is prime.
And here is another test (due to Shur) that requires no root testing.
Theorem $\ $ $\rm\:n>0,\,\ c_i\in\Bbb Z\:$ $\Rightarrow$ $\rm\ \dfrac{x^n}{n!} +\, c_{n-1}\dfrac{x^{n-1}}{(n-1)!} +\,\cdots + c_1 x \pm1 \ $ is irreducible in $\rm\,\Bbb Q[x].$
A: The rational root theorem can only help you find if there are rational
roots. 
A polynomial is irreducible $\implies$he doesn't have any roots.
A polynomial doesn't have any roots $\not\implies$ it is irreducible
.
However, the last implication is correct if the degree of the polynomial
is $\leq3$ since the factorization of such polynomial must have a
linear factor 
