f(x)=x^5 + 5*x^4 - 5*x^3 - 25*x^2 + 4*x + 20
By examining the coefficients of the polynomial, we find that
1+5-5-25+4+20=0 => (x-1) is a factor
Now, reverse the sign of coefficients of odd powers,
-1+5+5-25-4+20=0 => (x+1) is a factor
By the rational roots theorem, we can continue to try x=2, or factor x-2=0
2^5+5(2^4)-5(2^3)-25(2^2)+4(2)+20=0
and similarly f(-2)=0
So we have found four of the 5 real roots.
The remainder can be found by synthetic division as x=-5
Answer: The real roots of the given polynomial are: {-5,-2,-1.1.2}