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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.
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