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I am trying to find the partial fraction decomposition of $ (x^4+2)/(x^5+4x^3)$, which happens to be $(A/x) + (B/x^2) + C/(x^3) + ((Dx+E)/(x^2+6)$. My question is, why is there a " $+E$ " in the final term instead of being just $D$? How do I know when to put a $ + Constant $ in any of the terms?

Thanks!

Niko H
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    The fractions are of the form $\frac{p}{g^n}$ with $g$ irreducible in the coefficients that you are working on, and $p$ is of degree smaller than $g$. Supposedly you are working with coefficients such that $x^2+6$ doesn't factor (probably $\mathbb{Q}$ or $\mathbb{R}$). Then $g=x^+6$. The $p$ is some polynomial of degree less than $2$, so $1$ at most. – logarithm May 15 '19 at 21:11
  • To add to @logarithm (and +1 to him for his comment) : The $A,B,C$ are polynomials of degree $0$, and the $Dx + E$ is a polynomial of degree $1$ – MPW May 15 '19 at 21:12
  • So basically its because the denominator (g) is a 2nd degree polynomial, so the p must be a 1 degree polynomial at most? – Niko H May 15 '19 at 21:20
  • Yes, if $p$ had degree larger or equal to that of $g$ then you can divide $p=gb+r$, with $r$ this time having degree less than $g$ and then you could re-write $\frac{p}{g^n}=\frac{a}{g^{n-1}}+\frac{r}{g^n}$. Continuing dividing the $a$ by $g$ you can reduce to fractions in which the numerator always has degree less than that of $g$. – logarithm May 15 '19 at 21:28
  • But wait! You said the denominator. Note that it is not the denominator! The denominator is a power of some irreducible polynomial (one that doesn't factor). The degree of the numerator is less than the degree of that irreducible polynomial (which can be a lot less than the degree of the whole denominator). For example: $\frac{Ax^2+Dx+E}{(x^2+6)^{100}}$ is not yet a simple fraction in this sense. – logarithm May 15 '19 at 21:31
  • Thank you. But doesn't that mean that on the second term it should be $Bx+k$ instead of just $B$ since $x^2$ is 2nd degree. So $Bx+k$ has to be 1 degree at most? Same with the third term – Niko H May 15 '19 at 21:37
  • The answer is exactly my previous comment. In the fraction $\frac{E}{x^2}$, the $g(x)$ is $x$, not $x^2$, $n=2$ and $p$ must have degree less than that of $g$. Since $g$ has degree $1$, then $p$ must have degree $0$. – logarithm May 15 '19 at 21:42
  • Oh okay. Thank you! – Niko H May 15 '19 at 21:47

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