Algorithm to Remove Radicals From Denominators Based on the denominator s(y) generate a factor which when multiplied times the ratio will remove y from the denominator s(y). Let y be defined by a polynomial p(y)=0. Let s(y) be a polynomial in y with deg(s(y)) < deg(p(y)). Psuedo-dividing p(y) by s(y), the psuedo-quotient q1(y) and the psuedo-remainder r1(y) satisfy lc^(n-m+1)*p(y) = q1(y)*s(y) + r1(y) ; deg(r1(y)) < deg(s(y)) If deg(r1(y))=0 then q1(y)*s(y) = -r1 else psuedo-divide p(y) by r1(y) lc^(n-m+1)*p(y) = q2(y)*r1(y) + r2(y) ; deg(r2(y)) < deg(r1(y)) This is repeated until deg(rk(y)) = 0. Thus q1(y)*q2(y)*...*qk(y)*s(y) = (-1)^k * rk. The product q1(y)*q2(y)*...*qk(y) is a factor which when multiplied times a ratio will remove y from the denominator s(y). Ira Gessel suggests instead using the extended Euclidean Algorithm. Given p(y) and s(y) as above, it produces polynomials a(y) and b(y) such that a(y)*p(y) + b(y)*s(y) = gcd(p(y),s(y)) = r1. Because p(y)=0 the product b(y)*s(y) = r1. If rk=0 then p(y) and s(y) have a common factor and hence we are dividing by zero. An example of this is 1/(y+2) where y is defined by y^2 - 4 = 0. If deg(s(y))=1 the first algorithm terminates after the first division; The second algorithm will require more divisions.