Consider
Now the numerator has 4 zeroes. And the denominator has 6 zeroes? And and have triple zeroes in the denominator?
Thus has a double pole at ?
Is this correct?
I don't know what you mean by a "zero", but it is true that 0 is a zero of the numerator with multiplicity 4, and multiplicity 6 in the denominator.
If it is indeed a pole (you have to prove it by finding the Laurent series of this function in a neighbourhood of 0), it is likely to be a double pole...