# How to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial?

Rational function is defining as a polynomial with real coefficients over polynomial with real coefficents, how to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial since it is very troublesome?

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You may please want to clarify what an infinite discontinuity is. Does it mean that the function has a vertical asymptote at that point? – user21436 May 2 '12 at 21:16
@KannappanSampath It must be that case. – Pedro Tamaroff May 2 '12 at 21:18
Look at $P(x)/Q(x)$. There is discontinuity at $x=a$ precisely if $x-a$ is a factor of $Q(x)$, so to know the discontinuities, you will have to know the real $a$ such that $Q(a)=0$. This would give at least a partial factorization of $Q(x)$. To identify the removables can be easier, since the factor $x-a$ would have to occur in $Q(x)$ no more often than it does in $P(x)$. You can identify common factors by finding the greatest common divisor of $P(x)$ and $Q(x)$, which can be done by a variant of the Euclidean Algorithm. – André Nicolas May 2 '12 at 21:22
You can (approximately!) find the poles of a rational function without factoring, but the method I know requires a series expansion of your rational function, and I don't think that is any easier to do... – J. M. May 2 '12 at 21:22
@KannappanSampath Ok. You can delete the comment. – Pedro Tamaroff May 2 '12 at 21:25

Factoring is not needed, only much more efficient gcd computations. Suppose that your rational function $\rm\:F/G\:$ reduces to $\rm\:f/g\:$ in lowest terms, after cancelling out $\rm\:d = gcd(F,G).\:$ Then any root $\rm\:r\:$ of $\rm\:g\:$ is non-removable, since it cannot also be a root of $\rm\:f,\:$ else $\rm\:x-r\:$ divides both $\rm\:f,g\:$ contra $\rm\:gcd(f,g) = 1.\:$ So roots of $\rm\:d = gcd(F,G)\:$ not also roots of $\rm\:g\:$ are removed singularities, since $\rm\:F/G\:$ is singular at $\rm\:r\:$ by $\rm\:d(r)=0\:\Rightarrow\:F(r) = 0 = G(r),\:$ but $\rm\:f/g\:$ is not singular at $\rm\:r\:$ since $\rm\:g(r)\ne 0.\:$ These are precisely the roots whose numerator multiplicity is $\ge$ denominator multiplicity, so they are removed from the denominator when cancelling out $\rm\:gcd(F,G).$