How does one prove that a polynomial has no rational roots in general? How can we prove that a polynomial only has rational roots when we know the coefficients and the degree? For instance, in illustration, how would we show this for $x^8 +2x^7+3x^6+5x^5+4x^4+3x^3+2x^2+x-1$?
 A: Hint: By the Rational Root Theorem, the only rationals that could conceivably be roots are $\pm 1$. Plug in. Neither works. 
In general, if we have a polynomial $P(x)$ with integer coefficients, where $P(x)=a_0x^n+\cdots +a_n$, where $a_0\ne 0$, $a_n\ne 0$, then the only conceivable rational roots of $P(x)$ are of the form $\dfrac{a}{b}$, where $a$ is a divisor (possibly negative) of $a_n$ and $b$ is a positive divisor of $a_0$.
So there is a finite list of candidates. 
If we try them all, and nothing works, there are no rational roots. That's what happened in our concrete case.
To check whether there are any non-rational roots, find all the rational roots. 
We do have to check for multiple roots, so there is a need for some care. 
For example, we can use the Rational Root Theorem to show that the only rational root of $x^3-3x^2+3x-1$ is $1$. However, since our polynomial is $(x-1)^3$, the number $1$ is a triple root of the polynomial. 
If the sum of the number of rational roots (counting multiple roots according to their multiplicity) is not $n$, then there are (possibly non-real) roots that are not rational. 
Remark: The Rational Root Theorem is more useful in "made up" problems (homework, tests) than in "real life." Often in a problem, when you have say a cubic, and you need the roots, the numbers will have been chosen so that there is a reasonably simple rational root. The same is often true of quadratics in high school. Not so much later, since the Quadratic Formula gives a general procedure for solving quadratic equations.  
