When we perform a Partial Fraction Decomposition and one of the solutions of the denominator is a multiple solution (let's say quadratic), we write: $$\frac{A_{1}}{(x-x_{i})} + \frac{A_{2}}{(x-x_{i})^2}$$ My teacher said, that in the right denominator it is still a linear factor. But why is that (I mean it is of 2.degree) still a linear factor. It is more interesting with sums of these up to higher degrees, as in PFD your goal is to have linear and irreducible quadratic terms in the denominators of the partial fractions. So what about $\frac{A}{(x-x_{i})^3}$ or higher degrees of the denominator? Why are they still linear factors?
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1You might find this question and its answers helpful. – amd May 31 '16 at 17:31
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The multiplicity of the factor is greater than $1$, but the degree of the factor is still $1$. – Peter May 31 '16 at 17:38
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And why would that be? – Magnetar May 31 '16 at 18:14