Question 548737
I would first look for a rational zero and divide.
A rational zero would be a fraction whose numerator divides 580 and whose denominator divides 13.
The prime factorization of 580 is {{{580=2^2*5*29}}}
That means it has 12 positive divisors: 1,2,4,5,10,20,29,58,116,145,290, and 580.
Considering the possible denominators 1 and 13 (divisors of 13), that would make 24 positive rational numbers. With the corresponding 24 negative ones, that's 48 possible zeros to try.
Luckily, 2 is a zero of that polynomial. Dividing by {{{x-2}}}
I find {{{13x^3-185x^2+608x-580=(x-2)(13x^2-159x+290)}}}
Applying the quadratic formula to {{{13x^2-159x+290=0}}}
we can find any remaining zeros.
{{{x=(-(-159)+- sqrt((-159)^2-4*13*290))/(2*13)=(159+-sqrt(25281-15080))/26=(150+-sqrt(10201))/26=(159+-101)/26}}}
That gives us {{{x=10}}} and {{{x=29/13}}} as the last two zeros.