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Consider the two decompositions:

$\frac{x^2+1}{(x+2)(x-1)(x^2+x+1)}= \frac{A}{x+2} + \frac{B}{x-1} + \frac{Cx +D}{x^2+x+1}$,

and

$\frac{x^2+1}{(x-1)^3(x^2+1)^2}= \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{Dx +E}{x^2+1}+ \frac{Fx +G}{(x^2+1)^2}$

Can someone explain to me why both the

$\frac{Cx +D}{x^2+x+1}$ term in the first, and the

$\frac{Dx +E}{x^2+1}$ term in the second, get a numerator of the form $Ax+B$?

Lopey Tall
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    The decomposition theorem asserts that the numerator has degree less that the irreducible factor in the denominator. As the latter is a quadratic polynomial, the numerator has degree at most $1$ – Bernard Nov 22 '19 at 12:13
  • I see! So by this assertion, the numerator of a term with a denominator of the form $x^3+1$ (or $x^3+3x^2+x-2$ for that matter) would be $Ax^2+Bx+C$?? – Lopey Tall Nov 22 '19 at 13:07
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    Yes (except that over $\mathbf R$, no cubic polynomial is irreducible). – Bernard Nov 22 '19 at 13:09
  • Ooo good one! Thank you so much! Wouould you prefer I close the question, or select your future answer as the best? – Lopey Tall Nov 22 '19 at 13:18
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    As it maybe useful to others, I'll post an answer in a moment. – Bernard Nov 22 '19 at 13:25

2 Answers2

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The decomposition theorem over a field $F$ asserts that any rational fraction $\dfrac FG$ has a unique decomposition as described below:

Let $G=P_1^{n_1}\dotsb P_r^{n_r}$ be a decomposition of the denominator into irreducible factors, an $Q$ be the quotient in the Euclidean division of $F$ by $G$ . There exist a unique list of polynomials: $$A_{1,1},\dots A_{1,n_1},\dots, A_{r,1},\dots A_{r,n_r}$$ such that, for each $i=1,\dots,r$ , $\;\deg A_{i,1},\dots,\deg A_{i, n_i}<\deg P_i$, and $$\frac F G=P+\sum_{k=1}^{n_1}\frac{A_{1,k}}{P_1^k}+\dotsb+\sum_{k=1}^{n_r}\frac{A_{r,k}}{P_r^k}.$$

Now, if the field is $F=\mathbf R$, the only irreducible polynomials are either linear polynomials, and the corresponding coefficients $A_{i,k}$ have degree $0$, i.e. are constants, or quadratic polynomials with a negative discriminant, and the corresponding coefficients are polynomials of degree at most $1$.

Bernard
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$\dfrac{A}{(x+1)} + \dfrac{B}{(x+1)^2} + \dfrac{C}{(x+1)^3} = \dfrac{A(x+1)^2 + B(x+1)+C}{(x+1)^3} = \dfrac{A'x^2 + B'x+C'}{(x+1)^3}$

Steven Alexis Gregory
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