Partial fraction decomposition corresponding numerator question

Question

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 $Ax+B$?

Answer 1

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 $1$.

Answer 2

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