Discussion
A repeated linear factor of multiplicity two in the denominator contributes
Answer: D
A factor \( (x-a)^{2} \) needs \( A/(x-a) + B/(x-a)^{2} \), two constants for the two degrees of freedom lost. One term alone cannot represent a general numerator of degree one over the square. A linear numerator over a squared factor is the form reserved for an irreducible *quadratic* factor
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