The rational root theorem states that the rational roots of a polynomial can only be in the form p/q, where p divides the constant term, and q divides the leading term.
In your case, both the leading term 5 and the constant term 11 are primes, so their only divisors are 1 and themselves.
So, the only feasible solutions are
[tex]\pm\dfrac{11}{5},\quad\pm 11,\quad \pm\dfrac{1}{5}, \pm 1[/tex]
For the record, in this case, none of the feasible solutions are actually a root of the polynomial.
